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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?

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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

Nothing online seems 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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    $\begingroup$ I've deleted some comments which were close to answering the question. Everyone please keep in mind that comments are meant for suggesting improvements to, or requesting clarification on, their parent post, not for answering. $\endgroup$
    – David Z
    Dec 31, 2018 at 10:08

5 Answers 5

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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.

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    $\begingroup$ 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. $\endgroup$
    – InertialObserver
    Dec 30, 2018 at 23:40
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    $\begingroup$ 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. $\endgroup$
    – M. Winter
    Dec 31, 2018 at 1:46
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    $\begingroup$ In classical physics, we represent systems in terms of their degrees of freedom, as functions of time. It is very common to talk about the* state* of a system at a moment in time. That state is as close to a snapshot" as we can get. That kind of "snapshot" not only describes the configuration of a system, but also describes its momenta. In the case of the electromagnetic fields, the field itself has momentum. If you know the field (E and B) everywhere, you know the rate of change of the field.That is, the "momentum" of the field is part of the snapshot. $\endgroup$
    – S. McGrew
    Dec 31, 2018 at 4:49
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    $\begingroup$ In such a snapshot, it is reasonable to imagine that the fields are frozen and a probe particle is moved here and there to measure the fields. Remove the probe and "unfreeze" the fields, and the fields evolve just as if they had never been frozen. But we, who have the imaginary power to freeze and probe, can then use Maxwell's equations to predict accurately the fields' evolution. $\endgroup$
    – S. McGrew
    Dec 31, 2018 at 4:55
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    $\begingroup$ Electromagnetism is weird, innit? Entirely the result of relativity, despite the incredibly slow movement of the electrons...but still enough to cause length contraction and thereby a charge. $\endgroup$
    – Draco18s no longer trusts SE
    Dec 31, 2018 at 14:59
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Since you mention quantum phenomena, yes. In quantum mechanics the wave function carries all the information a particle can have, including momentum. You can see see this momentum by how much the phase rotates in space. If you calculate the expectation value of the momentum of a wave function that is flat, which has the same phase everywhere, you get zero. Any function that is real-valued is flat. You can give this wave function momentum by multiplying it by a rotating phase factor like $e^{ikx}$.

To make this more concrete I added an example from a quantum simulation I made some time ago. It simulates the one dimensional Schrodinger equation but it plots it in 3D so both the real and imaginary parts are drawn. From this you can immediately see that the bottom one has net momentum because the phase rotates like a corkscrew. Note that I made the first image static because it evolves in a really weird way which would distract from this question but in total it moves as much to the right as to the left.

So in conclusion, quantum mechanically you can see a difference between moving and static objects. A moving object has an overall rotating phase factor that makes the wave function look like a corkscrew. The faster it rotates in space, the greater the velocity of the object. In your case you have a large and 3D object which makes it harder to imagine but the same principle still holds.

Gaussian wave function Wave packet

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  • $\begingroup$ It seems to me that a flat wavefunction connected to a freely moving particle means that this freely moving particle has a momentum with zero uncertainty. In contrast, the position of the particle has infinite uncertainty: the flat wavefunction is spread out over the entire (position) space. Even though the wavefunction is flat, it has a single wavelength, corresponding to one value of its momentum. There is no superposition of different plane waves (like in your animation) which would give an uncertainty in both the particle's momentum and position. Besides, the asteroids are macroscopic. $\endgroup$
    – Deschele Schilder
    Dec 31, 2018 at 15:29
  • $\begingroup$ @descheleschilder I think I have caused some confusion with the unconventional use of the word flat. I mean any function that has a constant phase. So if you have a wave function that has only a real component the expectation value of the momentum will be zero. Example: if you take $\Psi(x)=Ae^{-x^2}$, which is real, the expectation value will be zero; $\langle{\Psi|}p|{\Psi}\rangle=0$. $\endgroup$
    – AccidentalTaylorExpansion
    Jan 1, 2019 at 21:27
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    $\begingroup$ @user3502079-First, I want to say sorry for my somewhat unpolite comment I made on your answer and my late reaction on your comment on it. But I see what you mean now. The quarter dropped! I see that by a flat wavefunction you mean a function that lies entirely in a plane, so the phase is indeed everywhere the same, in contrast to the moving (non-flat) function in your beautiful animation which has a "thickness" in the y-direction. So now I think it's a very nice answer and of course, I can't close this comment by wishing you a happy 2019! $\endgroup$
    – Deschele Schilder
    Jan 2, 2019 at 18:29
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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.

Suppose we know all of the properties of the asteroid in its own rest frame.Now, 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.

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  • $\begingroup$ 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? $\endgroup$
    – Jack
    Dec 30, 2018 at 22:59
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    $\begingroup$ 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. $\endgroup$
    – niels nielsen
    Dec 30, 2018 at 23:07
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    $\begingroup$ @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. $\endgroup$
    – InertialObserver
    Dec 30, 2018 at 23:25
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    $\begingroup$ When an object is moving relative to an observer it's length gets contracted by a factor of $\frac{1}{\gamma}$. When its at rest $\gamma =1$. I don't think I understand what youre saying. $\endgroup$
    – InertialObserver
    Dec 31, 2018 at 0:56
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    $\begingroup$ Okay---I'm not sure we need to belabor this, but for the record I was referring to your first paragraph, not your final paragraph with the explicit assumption. The OP asked, in essence, whether we'd see a difference between a photo taken while the asteroid is in motion and a photo taken after it stops (once again, all at once in the observer's frame). It seemed to me that (even if this was not your intention) your words could have misled a reader into believing that the two photos would show different lengths, which is a common misconception that it's better not to encourage. $\endgroup$
    – WillO
    Dec 31, 2018 at 4:00
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Let me approach this question qualitative in the "light" of the CMBR. In this article one can see that because of the movement of Earth relative to the CMBR a "dipole" develops with its axis parallel to the direction of the relative motion. Around one pole the temperature corresponding to the CMBR is higher than the temperature around the opposite pole (somewhat like an object moving through a gas experiences a higher pressure on the front side than on the rear side). This means that if an object is moving relative to the CMBR (not to be confused with the ether) the front side will obtain a slightly higher temperature than the rear side.

Applying this to the case of two asteroids, it is clear that there is a difference in the temperature gradients of both. There are two extremes:

  1. The asteroid that is dead still in space has no relative motion to the CMBR, so seen from this asteroid, the temperature gradient in the asteroid moving at $10 000(mph)$ is maximal.
  2. The asteroid that is dead still in space is moving at $10 000(mph)$ relative to the CMBR (this seems contradictory but according to special relativity one cannot say if an object is in absolute motion or not, which is why I wrote not to confuse the CMBR with the ether) so seen from this asteroid there is no temperature gradient in the other asteroid which isn't moving relative to the CMBR.

And of course, there is an infinite number of "in-between" cases. For example, if both asteroids move with the same (but opposite) speed relative to the CMBR, we have a symmetric situation and an observer on each asteroid sees the same temperature gradient (dipole) on the asteroid she looks at. Now I didn't do the calculation, but it is my guess that the temperature differences between the front and rear side of the asteroids are too tiny to measure, but in principle, they have to be there: the front side of an asteroid moving relative to the CMBR is in thermal equilibrium with the radiation of a black body that has a higher temperature than the black body with which the rear side is in equilibrium. It is a dynamical equilibrium though because heat is constantly moving from the front to the rear.

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The physical differences are many. The most obvious is that they are 2 different objects.
Let's assume they have identical mass distribution because they are composed of identical particles. Their location and orientation still differ.
Regardless of relativistic effects, one is described as still, the other in motion. Without specifying a single observer frame, I assume a majority of local observers would reach consensus on these labels. That makes it valid to consider those observers as data points in themselves. The gravity of the still asteroid will have affected the motion of local observers in a uniform way. It has a "sphere of influence".
The gravity of the moving asteroid will have affected the motion of a different set of local observers in a non-uniform way. It has an "egg of influence". The faster the asteroid is moving, the longer and thinner the "egg" becomes. The wake of the egg vs. the wake of the sphere will be discernible in a snapshot.

Rather than just looking at reality through the frame of one observer, we need to consider the perspectives and relative movements of all "eyewitness" observers, and just as important, observers nearby that "didn't see nuthin", and the pattern of the border between them.
Like a detective canvasing a crime scene to decipher what has happened; 2 observers may have wildly differing stories about what happened due to their perspective, equally valid, equally true, but there is a larger truth. The ground truth that satisfies all observations.

Captain Obvious wonders if this may be part of a more intuitive way to explain spacetime, time dilation, foreshortening. An object approaching the speed of light would have such a thin "egg" that it does not affect objects perpendicular to its path at all.

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