Special Relativity's effect on density

[MiLara]

Special relativity states that according to an observer at rest, a measuring stick on a moving platform will appear shorter.

Would this observer still see the measuring stick as comprising of the same amount of atoms as the observer who is at rest with respect to the measuring stick? If this is the case, would the first observer actually measure the measuring stick as being more dense since it's length is contracted?

I have a sense that this would violate conservation of mass and energy as this should still hold true regardless of the reference frame. I am no expert on relativity, so any insight as to where my logic is flawed would be greatly appreciated.

 

[phinds]

Special relativity states that according to an observer at rest, a measuring stick on a moving platform will appear shorter.

Would this observer still see the measuring stick as comprising of the same amount of atoms as the observer who is at rest with respect to the measuring stick? If this is the case, would the first observer actually measure the measuring stick as being more dense since it's length is contracted?

I have a sense that this would violate conservation of mass and energy as this should still hold true regardless of the reference frame. I am no expert on relativity, so any insight as to where my logic is flawed would be greatly appreciated.

Nothing is violated since the observer KNOWS that his observation is only valid in his reference frame and not in the object's rest frame where it counts.

 

[Dale]

ould this observer still see the measuring stick as comprising of the same amount of atoms as the observer who is at rest with respect to the measuring stick?

Yes.

would the first observer actually measure the measuring stick as being more dense since it's length is contracted

I am not sure what you mean by "first" observer since you only described one observer. However, in any case the direct answer is that the density of an object is higher in a frame where it is moving than in its rest frame.

have a sense that this would violate conservation of mass and energy as this should still hold true regardless of the reference frame

No conservation law is violated by the above. However, it is very important to understand the difference between "conserved" and "invariant". A conserved quantity does not change over time in a given reference frame. An invariant quantity is the same in all reference frames.

Energy is conserved, but not invariant. Different frames will have different values for the energy, but in each frame that value will not change over time.

Mass (the usual invariant mass used in modern relativity) is both conserved and invariant. Different frames will agree on the mass and also find that it will not change over time.

 

[MiLara]

Yes.

I am not sure what you mean by "first" observer since you only described one observer. However, in any case the direct answer is that the density of an object is higher in a frame where it is moving than in its rest frame.

No conservation law is violated by the above. However, it is very important to understand the difference between "conserved" and "invariant". A conserved quantity does not change over time in a given reference frame. An invariant quantity is the same in all reference frames.

Energy is conserved, but not invariant. Different frames will have different values for the energy, but in each frame that value will not change over time.

Mass (the usual invariant mass used in modern relativity) is both conserved and invariant. Different frames will agree on the mass and also find that it will not change over time.

If energy and mass are physically the same thing, How can one be invariant and conserved and the other just be conserved?
Also, is the increased relative density due to the kinetic energy of the measuring stick?

 

[weirdoguy]

If energy and mass are physically the same thing

They are not. You are probably thinking about rest energy, but that's not total energy.

 

[m4r35n357]

Special relativity states that according to an observer at rest, a measuring stick on a moving platform will appear shorter.

That is a simplification, and not a great one IMO.

The observer will "measure" (in some sense) the stick to be shorter, but the the ends of the stick as measured are not the same age in the stick's frame. Length contraction is not what it might seem at first! In any case none of this represents what you would really see, which is a bit more complicated . . . this video will give you a better idea.

 

[Bartolomeo]

That is a simplification, and not a great one IMO.

The observer will "measure" (in some sense) the stick to be shorter, but the the ends of the stick as measured are not the same age in the stick's frame. Length contraction is not what it might seem at first! In any case none of this represents what you would really see, which is a bit more complicated . . . this video will give you a better idea.

That means, if I know readings of clocks on the ends of the rod, I can determine it's velocity and direction of it's motion. Am I right?

 

[m4r35n357]

That means, if I know readings of clocks on the ends of the rod, I can determine it's velocity and direction of it's motion. Am I right?

I think so (together with the length you "measure"). If that is too vague, sorry but I don't really go in for this type of calculation as I hinted above.

 

[Comeback City]

If energy and mass are physically the same thing, How can one be invariant and conserved and the other just be conserved?
Also, is the increased relative density due to the kinetic energy of the measuring stick?

Einstein related Energy and Mass through e²=p²c²+m²c⁴
That doesn't mean they are physically the same thing, though.

 

[pervect]

Special relativity states that according to an observer at rest, a measuring stick on a moving platform will appear shorter.

Would this observer still see the measuring stick as comprising of the same amount of atoms as the observer who is at rest with respect to the measuring stick? If this is the case, would the first observer actually measure the measuring stick as being more dense since it's length is contracted?

I have a sense that this would violate conservation of mass and energy as this should still hold true regardless of the reference frame. I am no expert on relativity, so any insight as to where my logic is flawed would be greatly appreciated.

The usual formulation focuses on energy and not mass. The topic of mass in special relativity would probably require a separate post, I'll just briefely mention that it's worth learning about how "relativistic mass" is different from "invariant mass", and the importance of being clear about which concept of mass one is a) personally using and b) which concept the author of an article or paper or post on PF that one is reading is using. Confusion arises when the reader's concpets differ from the authors concept. To avoid a lengthly digression (such as which one is better), I'll focus on energy and it's conservation.

In special relativity, energy and momentum are both regarded as part of something larger, called the energy-momentum four vector. This can be regarded as being a consequence of the fundamental inter-relation between space and time. Note that length contraction can also be regarded as a consequence of this same relationship, so the two are closely related - and not particularly intuitive until one learns SR.

See the wiki article on the energy-momentum 4-vector <<link>>.

The density of energy/momentum is modeled by another mathematical object, called the stress-energy tensor. The wiki article is here <<link>>, but it might not make a lot of sense without the right backround. The stress-energy tensor can be regarded as describing the flow of energy-momentum.

If you read the details of the wiki article (or better yet a textbook reference), you'll see that there are applicable conservation laws, but the mathematical form of these laws and the mathematical entities (such as four-vectors and the stress-energy tensor, which is a rank 2 tensor) that are used to describe the applicable conservation laws may not be familiar. They're still there though.

The number of atoms in the bar does not changed, of course. The mathematical object that describes the density of atoms per unit volume is known as the number-flux four vector. There used to be a brief (and not very clear) description of it in the Wiki, but I don't see it anymore. This article <<link>> describes the number-flux four-vector and how it can be used to motivate the stress-energy tensor of a swarm of particles (for instance, a gas made up of moving atoms). But it's rather advanced.

The mathematical laws that describe the conservation of particles, and also the conservation of charge, are called the "continuity equation". An example for how the applicable laws look for the conservation of charge is given in this wiki article <<link>>. The same principles apply to the conservation of atoms.

For the stress-energy tensor, the applicable conservation law says that the divergence of the stress-energy tensor is zero. I'll give an honorable mention to the book "Div, Grad, Curl and all that" https://www.amazon.com/dp/0393925161/?tag=pfamazon01-20, though this book gives only the vector calculus version of the relationship between conservation laws and divergence free flows. This is probably a good thing though, the use of vector calculus rather than tensors make the concepts more accessible.

 

[Dale]

If energy and mass are physically the same thing, How can one be invariant and conserved and the other just be conserved?

There is a concept called relativistic mass, which is physically the same thing as energy. It has fallen out of use, and modern physicists use the concept of invariant mass now. The invariant mass is physically different from energy.

is the increased relative density due to the kinetic energy of the measuring stick

The increased mass density is purely due to length contraction. The increased energy density would be due to both length contraction and also the increased KE

 

[Jeronimus]

That means, if I know readings of clocks on the ends of the rod, I can determine it's velocity and direction of it's motion. Am I right?

You could determine the rod's speed, yes. If you placed clocks along the rod which are at sync in the rod's rest frame, you could then in theory determine its exact speed by checking the difference in clock counts between clocks on the rod from any given inertial frame of reference.
From the diagrams below, you can see that the difference in time between two clocks at the endpoints of a rod, having the size of 5 lightseconds, would be 2.5 seconds if those clocks were to be in sync in the rod's rest frame.
Which you could then use to calculate the relative speed between you and the rod. 0.5c in this case.The direction, i don't think so.

This is how it would look like for a rod with a length of 5 lightseconds in the left x-t diagram, when observed by someone who is moving at 0.5c relative to the rod.

In the left diagram, the red line on the x-axis represents a rod with 6 clocks on top of the rod, all synced with a clock count of 0 seconds. Those clocks are all on top of the x-axis (simultaneous)

Those 6 clocks with a clock count of 0 seconds are not synced anymore when observed by an observer who is moving relative to the rod ( v=0.5c in the case of the right diagram).

The diagonal red line in the right diagram is where those 6 clocks with a clock count of 0 are on. Their t-position is not equal anymore.
Let's call the 6 clocks with a clock count of 0, instances of those 6 clocks, which lie on the worldlines of those 6 clocks.

So if we were to define a rod by being composed of the same _instances_ of atoms in both frames, we would be looking at the red lines in both cases.

However, that is not how we define the length of an object, or the object itself for that matter. To measure the length of an object, we measure two endpoints of the object having the same t- position (are simultaneous within any given inertial frame of reference).

In the case of the right diagram, this would be the orange line.

This orange line IS the rod by definition, and is composed of the "same" atoms by definition. Except, those atoms are different instances of the atoms which are either older or younger(compared to the "rest frame rod"), depending on the velocity vector.

The orange line in the right diagram, representing the moving rod is only about 4.3 lightseconds in size, compared to the "same" rod in the left diagram, which is measured to be 5 lightseconds. Yet, both have the same amount of atoms between the endpoints, just as they have the same amount of clocks fitting between them. 6 in this case.

They are different instances(older or younger) of the "same" clocks, with their worldlines (red and pink in the right diagram) all crossing through the orange line representing the rod.

edit: Maybe someone can formulate it better. It's not really easy to pack this into words. - I tried :D

 

[madScientist404]

I would just say, according to that particular observer it is space itself that is contracted. Therefore everything is contracted, also the atoms in the stick. In both situations the stick contains the same amount of atoms with the same direction relative to each other and therefore there is no change in density.

 

[Ibix]

I would just say, according to that particular observer it is space itself that is contracted.

Not really. Rather, different observers use different definitions of space, which intersect the worldtubes of objects in different ways. So the atom count doesn't change, as you say, but it is reasonable to say that density is a frame-dependent quantity.

 

[Battlemage!]

Not really. Rather, different observers use different definitions of space, which intersect the worldtubes of objects in different ways. So the atom count doesn't change, as you say, but it is reasonable to say that density is a frame-dependent quantity.

Does this analogy apply to time? That is, all types of clocks run at different rates for the two inertial observers, so does it make sense to say time "dialates?" Time dilates, but only lengths contract? (rather than space contracts)

 

[Ibix]

@Battlemage! - I'm fine with "length contraction". A moving ruler is shorter than a stationary one. Two things that are 1m apart in their shared rest frame are $1/\gamma$ apart in another. But "space contraction" kind of implies you're doing something to spacetime, rather than just changing coordinates.

I think the reason that "time dilation" is ok is that the word "time" is doing double duty (maybe it should get overtime ) as both a component of spacetime and as measurements made in that direction. But we have two separate words for space and measurements of separation in space, and we shouldn't confuse them.

I must admit I hadn't thought about that until you asked, and I reserve the right to change my mind without notice.

 

[madScientist404]

if the observer moving with the stick measures the mass density as (rest mass)/(rest Volume) = m0/V0
and the stationary observer measures the mass density as (relativistic mass)/(contracted Volume) = m/V = γm0/V0/γ = γ^2 m0/V0
then mass density is (indeed) frame dependent.
In my mind the two γ canceled They do not, which makes my earlier statement false.

 

[weirdoguy]

relativistic mass

Maybe not:
https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

 

[madScientist404]

i know, but i could not come up with a better name for m.