If time and space are relative, is temperature relative too?
December 6, 2006 8:51 PM Subscribe
I’ve only browsed A Brief History of Time, but in the bathtub the other day having a decent soak , I had a few overtired ideas I’d like to bounce off someone with a knowledge of the field of physics and the theory of relativity… (more after the jump)
If time is relative to velocity, and time goes more slowly aboard an object going very fast vs one going very slow, and temperature is a measure of the average velocity/energy of particles, then does that mean that temperature is relative? And to someone stationary, that the temperature of a theoretical object going the speed of light would be absolute zero?
To take that one step futher, imagine an unreversible chemical reaction that occours at a certain temperature, and that that chemical system is aboard a spacecraft approaching the speed of light. Lets say to an observer on earth, the object never clears the threshold of reaction because as it heats, it is getting faster and faster..
But on board the spaceship, the temperature does increase so that the reaction occours. Wouldn’t that lead to some deviation of reality between an observer on earth and one on the ship?
It all sounds really stoned I know, but I’d love to know more.
If time is relative to velocity, and time goes more slowly aboard an object going very fast vs one going very slow, and temperature is a measure of the average velocity/energy of particles, then does that mean that temperature is relative? And to someone stationary, that the temperature of a theoretical object going the speed of light would be absolute zero?
To take that one step futher, imagine an unreversible chemical reaction that occours at a certain temperature, and that that chemical system is aboard a spacecraft approaching the speed of light. Lets say to an observer on earth, the object never clears the threshold of reaction because as it heats, it is getting faster and faster..
But on board the spaceship, the temperature does increase so that the reaction occours. Wouldn’t that lead to some deviation of reality between an observer on earth and one on the ship?
It all sounds really stoned I know, but I’d love to know more.
The step where I don't quite follow you is where you say "imagine an unreversible chemical reaction that occours at a certain temperature, and that that chemical system is aboard a spacecraft approaching the speed of light." If you're imagining, say, that we have someone slowly turning up the temperature until at some point the reaction occurs, then there's no real paradox here: it just takes more time for the heater to get up to heat because of the relativistic time dilation, and there's no more paradox in saying "the reaction never occurs on a spaceship travelling at the speed of light" than there is in saying "clocks on board a ship travelling at the speed of light appear to stop."
posted by Johnny Assay at 9:21 PM on December 6, 2006
posted by Johnny Assay at 9:21 PM on December 6, 2006
No, I think what he's saying is that the chemicals are on the ship, at some constant temperature in that reference frame, below the reaction temperature. An observer in a different reference frame, potentially seeing the chemicals as being at a higher temperature, might expect the reaction to have taken place.
This paradox would seem to suggest that temperature is Lorentz-invariant.
posted by kickingtheground at 9:28 PM on December 6, 2006
This paradox would seem to suggest that temperature is Lorentz-invariant.
posted by kickingtheground at 9:28 PM on December 6, 2006
Response by poster: I was imagining a scenario where the chemicals are slowly heated at the same time the ship's velocity increases, at some rate correlating so that from the ground, the temperature is constant, and remains lower than the threshold..
Really my question is more about whether there is any scenario where a "relative" propery of temperature would cause some deveiation in reality based on observation point.
posted by upc_head at 9:40 PM on December 6, 2006
Really my question is more about whether there is any scenario where a "relative" propery of temperature would cause some deveiation in reality based on observation point.
posted by upc_head at 9:40 PM on December 6, 2006
Weight is relative - the faster an object travels, the lighter the object is. Light itself is weightless because it travels at the speed of light. If it were possible to capture light so that it does not move, it would be infinitely heavy.
However, I'm not sure if temperature is relative. If an object is colder the faster it is, then why is light itself not ice cold? Wouldn't that suggest that lasers would be freezing, not burning?
posted by pikaboy202 at 9:44 PM on December 6, 2006
However, I'm not sure if temperature is relative. If an object is colder the faster it is, then why is light itself not ice cold? Wouldn't that suggest that lasers would be freezing, not burning?
posted by pikaboy202 at 9:44 PM on December 6, 2006
Temperature is a measure of the random motion of molecules, so wouldn't their "speed" average to zero?
posted by weapons-grade pandemonium at 10:03 PM on December 6, 2006
posted by weapons-grade pandemonium at 10:03 PM on December 6, 2006
Imagine instead of temperature, a macroscopic system like balls bouncing around a basket on a spaceship flying away from the Earth. If they are going fast enough, they fly out of the basket. This is analogous to an irreversible chemical process that occurs at a certain temperature.
An Earthbound observer would see balls moving more slowly, but if he calculated their velocities accounting for the lorentz contraction, he'd be able to predict whether or not they were energetic enough to leave the basket.
posted by justkevin at 10:07 PM on December 6, 2006
An Earthbound observer would see balls moving more slowly, but if he calculated their velocities accounting for the lorentz contraction, he'd be able to predict whether or not they were energetic enough to leave the basket.
posted by justkevin at 10:07 PM on December 6, 2006
At first glance, relativistic temperature seems reasonable. The average kinetic energy of a set of particles on a speeding spaceship would be lower for an observer on earth than for an observer in the spaceship, so the temperature measured temperature would be lower for the earth-based observer.
However, no deviation in reality would occur in the situation presented. Suppose the observer on earth measures the temperature of the reaction container in the speeding spaceship and finds it is below the reaction temperature. Being a good scientist, the observer then measures the relative velocity of the spaceship, calculates the relativistic correction to the temperature and decides if the reaction will or will not occur. Her conclusion will match any observation made on board the spaceship during its flight or when it returns to earth.
posted by ssg at 10:42 PM on December 6, 2006
However, no deviation in reality would occur in the situation presented. Suppose the observer on earth measures the temperature of the reaction container in the speeding spaceship and finds it is below the reaction temperature. Being a good scientist, the observer then measures the relative velocity of the spaceship, calculates the relativistic correction to the temperature and decides if the reaction will or will not occur. Her conclusion will match any observation made on board the spaceship during its flight or when it returns to earth.
posted by ssg at 10:42 PM on December 6, 2006
Best answer: You're assuming simultaneity, and under relativity there is no simultaneity.
The effect you're postulating is one where different observers see different things which seem to contradict one another. But there are a lot of such things (e.g. the barn door paradox) and it turns out that all of those paradoxes are resolvable once one really understands the fact that there is no simultaneity between places which are physically separated.
The effects described by Special Relativity only make sense in inertial frames of reference. As soon as the frame of reference accelerates, all the rules change. (For instance, C isn't necessarily constant in an accelerating frame of reference.)
So you have two inertial frames of reference moving relative to one another at speeds approaching C. They fly past one another and that is the only instant when their time scales can be accurately referenced to one another.
As they fly apart, it makes no sense to say "When person A in frame of reference A sees thus-and-so, what does person B in frame of reference B see?" There is no time point in B's frame of reference which is "at the same time" as the moment when A made that observation.
The only way the two of them can again reference their time scales relative to one another is if they encounter each other again and that can only happen if one or both accelerate.
That's the explanation for the "twins paradox". One twin stays at home. The other one spends a lot of time in a space ship travelling near the speed of light. Eventually he returns. Each one thought he was standing still, and thought the other was moving fast. Each one should see the other as younger, right?
Nope! The one in the space ship had to accelerate in order to make a round trip, so it's him that suffers from time compression. It will unambiguously be the one in the space ship who is younger.
But what if his ship doesn't accelerate once he gets up to speed? Then he cannot come home, and since there is no way to reference any given point of time in the space ship with a point of time on the home planet, it doesn't make any sense to say that one of them is aging faster than the other. Their timescales are unreferenced. There's no simultaneity in frames of reference which are physically separated.
That's also the answer to your problem. If the two frames of reference are both inertial (i.e. neither accelerates) then observers in the two (one using a very powerful telescope) will see different things. Which one is right? They both are. They see different things because their time scales are unreferenced. The only way their time scales can become referenced is if they come into contact before the experiment begins and again after it and that requires one or both of them to accelerate. Who accelerates, and how much, and when, turns out to control what they'll discover once their time scales again become referenced.
The math works beautifully. The answers are unambiguous. But it's one of many counter-intuitive consequences of relativity. The assumption of simultaneity is a deep one that most people don't even realize they're making, and if you do assume the existence of simultaneity, then relativity won't make any sense for you.
posted by Steven C. Den Beste at 10:42 PM on December 6, 2006 [3 favorites]
The effect you're postulating is one where different observers see different things which seem to contradict one another. But there are a lot of such things (e.g. the barn door paradox) and it turns out that all of those paradoxes are resolvable once one really understands the fact that there is no simultaneity between places which are physically separated.
The effects described by Special Relativity only make sense in inertial frames of reference. As soon as the frame of reference accelerates, all the rules change. (For instance, C isn't necessarily constant in an accelerating frame of reference.)
So you have two inertial frames of reference moving relative to one another at speeds approaching C. They fly past one another and that is the only instant when their time scales can be accurately referenced to one another.
As they fly apart, it makes no sense to say "When person A in frame of reference A sees thus-and-so, what does person B in frame of reference B see?" There is no time point in B's frame of reference which is "at the same time" as the moment when A made that observation.
The only way the two of them can again reference their time scales relative to one another is if they encounter each other again and that can only happen if one or both accelerate.
That's the explanation for the "twins paradox". One twin stays at home. The other one spends a lot of time in a space ship travelling near the speed of light. Eventually he returns. Each one thought he was standing still, and thought the other was moving fast. Each one should see the other as younger, right?
Nope! The one in the space ship had to accelerate in order to make a round trip, so it's him that suffers from time compression. It will unambiguously be the one in the space ship who is younger.
But what if his ship doesn't accelerate once he gets up to speed? Then he cannot come home, and since there is no way to reference any given point of time in the space ship with a point of time on the home planet, it doesn't make any sense to say that one of them is aging faster than the other. Their timescales are unreferenced. There's no simultaneity in frames of reference which are physically separated.
That's also the answer to your problem. If the two frames of reference are both inertial (i.e. neither accelerates) then observers in the two (one using a very powerful telescope) will see different things. Which one is right? They both are. They see different things because their time scales are unreferenced. The only way their time scales can become referenced is if they come into contact before the experiment begins and again after it and that requires one or both of them to accelerate. Who accelerates, and how much, and when, turns out to control what they'll discover once their time scales again become referenced.
The math works beautifully. The answers are unambiguous. But it's one of many counter-intuitive consequences of relativity. The assumption of simultaneity is a deep one that most people don't even realize they're making, and if you do assume the existence of simultaneity, then relativity won't make any sense for you.
posted by Steven C. Den Beste at 10:42 PM on December 6, 2006 [3 favorites]
I was imagining a scenario where the chemicals are slowly heated at the same time the ship's velocity increases, at some rate correlating so that from the ground, the temperature is constant, and remains lower than the threshold.
If the ship's velocity is increasing, then that means it is accelerating. Measurements taken in an accelerating frame of reference are not reliable, and won't match predictions from Special Relativity.
posted by Steven C. Den Beste at 10:53 PM on December 6, 2006
If the ship's velocity is increasing, then that means it is accelerating. Measurements taken in an accelerating frame of reference are not reliable, and won't match predictions from Special Relativity.
posted by Steven C. Den Beste at 10:53 PM on December 6, 2006
Weight is relative - the faster an object travels, the lighter the object is
'Fraid you've got that backwards, picabo202
But on board the spaceship, the temperature does increase so that the reaction occours. Wouldn’t that lead to some deviation of reality between an observer on earth and one on the ship?
That surely is one of the most interesting results of relativity. Reality - what one observes- is not the same in different frames of reference
posted by Neiltupper at 11:16 PM on December 6, 2006
Here is the problem: You have to really understand that time is not absolute i.e., there is not a "blanket" of time that the whole universe can relate to. There is no "right now", only a "right here/now". This is related to the idea that time is just one of our dimensions, like the way we think of space-- also the reason we refer to the "space-time continuum".
So, if some people are watching chemicals react in a very fast spaceship (remember, those chemicals are moving at a normal speed relative to the people on the spaceship), they are going to react normally, the way they usually do.
Back on Earth, we see the spaceship going away, looking cooler all the time, redshifting, and certainly, if the spaceship is approaching the speed of light, the reaction is never happening. The people on the ship aren't aging. In fact, the whole spaceship doesn't seem to be doing a damn thing. From our observational standpoint, we are never going to get the information that the reaction (or anything, for that matter)has happened. But that IS the reality. That is exactly the same information that left the spaceship when the reaction was occurring and traveled through the spacetime fabric to come to us. Information is "reality", and it travels at the speed of light through the spacetime continuum, which warps and moves according to the movement (or energy/mass) of the objects in it.
posted by simonemarie at 11:22 PM on December 6, 2006
So, if some people are watching chemicals react in a very fast spaceship (remember, those chemicals are moving at a normal speed relative to the people on the spaceship), they are going to react normally, the way they usually do.
Back on Earth, we see the spaceship going away, looking cooler all the time, redshifting, and certainly, if the spaceship is approaching the speed of light, the reaction is never happening. The people on the ship aren't aging. In fact, the whole spaceship doesn't seem to be doing a damn thing. From our observational standpoint, we are never going to get the information that the reaction (or anything, for that matter)has happened. But that IS the reality. That is exactly the same information that left the spaceship when the reaction was occurring and traveled through the spacetime fabric to come to us. Information is "reality", and it travels at the speed of light through the spacetime continuum, which warps and moves according to the movement (or energy/mass) of the objects in it.
posted by simonemarie at 11:22 PM on December 6, 2006
I think the statement that the temperature changes as one moves closer to the speed of light is false.
posted by number9dream at 11:28 PM on December 6, 2006
posted by number9dream at 11:28 PM on December 6, 2006
number9dream, the temperature to an observer would appear to be cooler as one moves toward the speed of light, because objects would appear to be moving slower. The temperature on the ship would look the same to them.
Once they hit the speed of light and their time stops (from the reference of the observer), they would be at absolute zero. At the "same time" (which is the statement that is confusing everyone in the first place) whoever is in the spaceship could be curled up in a blanket reading a book.
posted by simonemarie at 11:36 PM on December 6, 2006
Once they hit the speed of light and their time stops (from the reference of the observer), they would be at absolute zero. At the "same time" (which is the statement that is confusing everyone in the first place) whoever is in the spaceship could be curled up in a blanket reading a book.
posted by simonemarie at 11:36 PM on December 6, 2006
By contact, do you mean that the objects are within distances and accelerations that are reasonably approximated by newtonian mechancs?
All of the measurement comparisons being discussed here are subject to latency due to speed-of-light delay. So referencing the time scales of two inertial frames of reference requires them to be close enough together physically so that errors induced by latency due to speed-of-light are negligible within the context of the kinds of measurements that are being made.
The proposed experiment doesn't work if both frames of reference are inertial. That means that the relative velocity is constant, which either means they're both accelerating the same amount in the same direction, or it means neither of them is accelerating at all. In either case, relativistic effects are not changing over the course of the experiment.
The only thing that is changing is the distance between them, and therefore the amount of time latency involved for observations by the guy with the telescope.
That means you can't get a situation where one guy thinks the ice has melted and the other guy thinks that it's too cold and will never melt. Either they'll both see it melt or neither will. But they'll disagree about how long it takes.
Each of them thinks he's standing still and the other is moving. Each of them thinks the other is subject to relativistic effects, in terms of changes in time, mass, and length.
Each of them will do the calculations based on the assumption that he is standing still and the other guy is moving. And each of them will calculate that the other is subject to an amount of relativity-induced error which explains the discrepancy.
When I saw how the math worked out, it made me start giggling in glee. It's really marvelous.
See, the thing is that in order to make the calculation, each of them has to assume simultaneity. But they'll calculate the time-point-of-simultaneity differently. There is one time that's referenced between the two frames, and each will calculate time durations before and after that based on the assumption that the other is moving.
"When it's 2:30PM for me, it's 1:30PM for him, because his time is compressed by his velocity." They'll both say that. And that means they're cross-tracking their respective time scales differently.
In actual fact, there is no correlation at all. The only point in time where "simultaneous" made physical sense was when the two were next to one another and they referenced their time scales.
So even though their observations disagree, in once sense they're both right. In another sense they're both wrong.
But the real point is that it isn't even meaningful to ask which is right. What each of them sees is consistent and agrees with the predictions made by Special Relativity, and that's the only thing that really matters. The discrepancy is due to the fact that there is no simultaneity.
The discrepancy only has to become physically resolved if they twice come near enough to reference their time scales, and for that to happen acceleration must occur, which invalidates the measurements made by one or the other (or both), and calculations about those measurements, in ways which are well understood.
posted by Steven C. Den Beste at 11:43 PM on December 6, 2006
All of the measurement comparisons being discussed here are subject to latency due to speed-of-light delay. So referencing the time scales of two inertial frames of reference requires them to be close enough together physically so that errors induced by latency due to speed-of-light are negligible within the context of the kinds of measurements that are being made.
The proposed experiment doesn't work if both frames of reference are inertial. That means that the relative velocity is constant, which either means they're both accelerating the same amount in the same direction, or it means neither of them is accelerating at all. In either case, relativistic effects are not changing over the course of the experiment.
The only thing that is changing is the distance between them, and therefore the amount of time latency involved for observations by the guy with the telescope.
That means you can't get a situation where one guy thinks the ice has melted and the other guy thinks that it's too cold and will never melt. Either they'll both see it melt or neither will. But they'll disagree about how long it takes.
Each of them thinks he's standing still and the other is moving. Each of them thinks the other is subject to relativistic effects, in terms of changes in time, mass, and length.
Each of them will do the calculations based on the assumption that he is standing still and the other guy is moving. And each of them will calculate that the other is subject to an amount of relativity-induced error which explains the discrepancy.
When I saw how the math worked out, it made me start giggling in glee. It's really marvelous.
See, the thing is that in order to make the calculation, each of them has to assume simultaneity. But they'll calculate the time-point-of-simultaneity differently. There is one time that's referenced between the two frames, and each will calculate time durations before and after that based on the assumption that the other is moving.
"When it's 2:30PM for me, it's 1:30PM for him, because his time is compressed by his velocity." They'll both say that. And that means they're cross-tracking their respective time scales differently.
In actual fact, there is no correlation at all. The only point in time where "simultaneous" made physical sense was when the two were next to one another and they referenced their time scales.
So even though their observations disagree, in once sense they're both right. In another sense they're both wrong.
But the real point is that it isn't even meaningful to ask which is right. What each of them sees is consistent and agrees with the predictions made by Special Relativity, and that's the only thing that really matters. The discrepancy is due to the fact that there is no simultaneity.
The discrepancy only has to become physically resolved if they twice come near enough to reference their time scales, and for that to happen acceleration must occur, which invalidates the measurements made by one or the other (or both), and calculations about those measurements, in ways which are well understood.
posted by Steven C. Den Beste at 11:43 PM on December 6, 2006
Sorry, a small correction: If two frames of reference are physically separated and both are accelerating the same amount in the same direction, they are still not inertial frames of reference.
posted by Steven C. Den Beste at 11:49 PM on December 6, 2006
posted by Steven C. Den Beste at 11:49 PM on December 6, 2006
Thanks, Steven, for writing an excellent explanation. (Now I don't have to try to bumble through one).
For those who are interested, I've got a favorite book on special relativity that I like to share with anyone who wants to see "the math work out"-- it's just algebra in this case (no calculus needed!). And it is truly beautiful (says the biased physicist).
The book is called "Space and Time in Special Relativity", by N. David Mermin. It's a wonderful little pink paperback book, and pretty readable. OP, wander over to the nearest university library and check it out-- I think it'll answer your relativity questions better than any of us can in a short askme thread.
Good luck!
posted by nat at 11:58 PM on December 6, 2006
For those who are interested, I've got a favorite book on special relativity that I like to share with anyone who wants to see "the math work out"-- it's just algebra in this case (no calculus needed!). And it is truly beautiful (says the biased physicist).
The book is called "Space and Time in Special Relativity", by N. David Mermin. It's a wonderful little pink paperback book, and pretty readable. OP, wander over to the nearest university library and check it out-- I think it'll answer your relativity questions better than any of us can in a short askme thread.
Good luck!
posted by nat at 11:58 PM on December 6, 2006
An important point: velocity is relative, but acceleration is not relative. The quantity of acceleration can be measured differently by different observers, but all observers will agree that acceleration is taking place, and agree about who is doing the accelerating.
I was imagining a scenario where the chemicals are slowly heated at the same time the ship's velocity increases, at some rate correlating so that from the ground, the temperature is constant, and remains lower than the threshold.
Both observers will unambiguously agree that the ship is accelerating. That means that measurements taken by the guy in the accelerating frame of reference are not reliable, and both of them will know it.
posted by Steven C. Den Beste at 12:22 AM on December 7, 2006
I was imagining a scenario where the chemicals are slowly heated at the same time the ship's velocity increases, at some rate correlating so that from the ground, the temperature is constant, and remains lower than the threshold.
Both observers will unambiguously agree that the ship is accelerating. That means that measurements taken by the guy in the accelerating frame of reference are not reliable, and both of them will know it.
posted by Steven C. Den Beste at 12:22 AM on December 7, 2006
Steven, could you please explain the case where a match is travelling vertically (in a horizontal orientation) at a low speed while its open matchbox is travelling horizontally (in a horizontal orientation) at a near-c speed. These movements are timed so that the match meets the matchbox at same time and place. Does the match fit in the matchbox given they both witness length contractions? Thanks.
posted by zaebiz at 1:25 AM on December 7, 2006
posted by zaebiz at 1:25 AM on December 7, 2006
The temperature isn't fundamental. It's a way of expressing the energy the particles in your flask of reactants have. The chemically important thing is how hard the chemicals bang into each other and whether that energy is enough to make the electrons in the molecules do whatever is necessary for the reaction. It doesn't matter what the ship is doing. It's the velocities of the reactants relative to each other that is important and this won't depend on the overall velocity of your spaceship.
zaebiz: look up the pole-barn paradox.
posted by edd at 1:37 AM on December 7, 2006
zaebiz: look up the pole-barn paradox.
posted by edd at 1:37 AM on December 7, 2006
edd: "zaebiz: look up the pole-barn paradox."
I know of this paradox. Also know as the ladder paradox. However I believe it is quite different. The "barn doors" as it were in the example I proposed, are always closed so there is not the same opportunity for differences in observations of simultaneity. I'm guessing it has something to do with the changing angle of the contraction axis.
posted by zaebiz at 1:47 AM on December 7, 2006
I know of this paradox. Also know as the ladder paradox. However I believe it is quite different. The "barn doors" as it were in the example I proposed, are always closed so there is not the same opportunity for differences in observations of simultaneity. I'm guessing it has something to do with the changing angle of the contraction axis.
posted by zaebiz at 1:47 AM on December 7, 2006
Late to the party. Lots of answers here. I'm certainly not the brightest lamp on the MetaFilter string, but this is how I think about it.
I'll try to keep this simple since it means I won't have to embarrass myself with my bad math. I don't really think we need it anyway. This explanation may be wordy but I hope it's understandable -- maybe if there are any physicists reading this they'll do me the favor of telling me if I'm right.
To answer the second question first, if an irreversible reaction occurs in one frame the other observer will see it happen also. They might not agree on when it happens. But they have to both see it happen eventually. There's no shortage of thought experiments we can do to convince ourselves of this (or at least myself - hopefully I'm right.) Let's just accept that for now.
Now why is that, if we measure temperature (of say, a gas) as a function of velocity? It's true that we'll see a gas in motion as having a lower r.m.s. velocity than an observer at rest in the gas's frame will. However, if we arrange a window on the spaceship (did I mention the experiment's taking place on a spaceship?) so that we can read Spaceman Spiff's thermometer when he passes us, we'll see his thermometer reading is sufficient to let the reaction occur -- in fact his thermometer reading is the same as it would be if we were doing the experiment on our coffee table (did I mention we're in our living room?)
The thing is, relativistic effects not only apply to the particles' velocity but also to their mass. And they don't only affect the gas but also the thermometer (or, equivalently, the other reaction components.) When you do the math it works out that the energy of the gas we observe from our living room chair is sufficient to cause the thermometer we see from our living room chair to read the temperature it reads.
**Lots of hand waving**
It's a little late here on the East Coast (at least for somebody my age) to do the math, but I'm pretty sure when I sit down tomorrow and actually do this that's how it's going to work out. One of the MetaBrains can probably work this out in their head and confirm or deny that this makes sense.
posted by Opposite George at 1:56 AM on December 7, 2006
I'll try to keep this simple since it means I won't have to embarrass myself with my bad math. I don't really think we need it anyway. This explanation may be wordy but I hope it's understandable -- maybe if there are any physicists reading this they'll do me the favor of telling me if I'm right.
To answer the second question first, if an irreversible reaction occurs in one frame the other observer will see it happen also. They might not agree on when it happens. But they have to both see it happen eventually. There's no shortage of thought experiments we can do to convince ourselves of this (or at least myself - hopefully I'm right.) Let's just accept that for now.
Now why is that, if we measure temperature (of say, a gas) as a function of velocity? It's true that we'll see a gas in motion as having a lower r.m.s. velocity than an observer at rest in the gas's frame will. However, if we arrange a window on the spaceship (did I mention the experiment's taking place on a spaceship?) so that we can read Spaceman Spiff's thermometer when he passes us, we'll see his thermometer reading is sufficient to let the reaction occur -- in fact his thermometer reading is the same as it would be if we were doing the experiment on our coffee table (did I mention we're in our living room?)
The thing is, relativistic effects not only apply to the particles' velocity but also to their mass. And they don't only affect the gas but also the thermometer (or, equivalently, the other reaction components.) When you do the math it works out that the energy of the gas we observe from our living room chair is sufficient to cause the thermometer we see from our living room chair to read the temperature it reads.
**Lots of hand waving**
It's a little late here on the East Coast (at least for somebody my age) to do the math, but I'm pretty sure when I sit down tomorrow and actually do this that's how it's going to work out. One of the MetaBrains can probably work this out in their head and confirm or deny that this makes sense.
posted by Opposite George at 1:56 AM on December 7, 2006
When you do the math it works out that the energy of the gas we observe from our living room chair is sufficient to cause the thermometer we see from our living room chair to read the temperature it reads.
That is, we observe the particles on the spaceship as being energetic enough to make the thermometer on the spaceship read a sufficient temperature for the reaction to occur (forgive me; it's late.)
posted by Opposite George at 2:00 AM on December 7, 2006
That is, we observe the particles on the spaceship as being energetic enough to make the thermometer on the spaceship read a sufficient temperature for the reaction to occur (forgive me; it's late.)
posted by Opposite George at 2:00 AM on December 7, 2006
Zaebiz, remember this that I wrote earlier?
So referencing the time scales of two inertial frames of reference requires them to be close enough together physically so that errors induced by latency due to speed-of-light are negligible within the context of the kinds of measurements that are being made.
I purposefully didn't give an absolute distance, because it really is a function of the measurements being made.
The two ends of the match are far enough apart, as a function of the kinds of measurements being made here, so that there's a non-negligible error in measurements between the two.
These movements are timed so that the match meets the matchbox at same time and place.
It isn't the "same time and place". It's two different places whose times can't be correlated to one another. Observers at each end of the match will observe somewhat different things. What each sees will be completely consistent, but they may disagree with one another.
(In extreme cases you can't even necessarily assume simultaneity over sub-atomic distances.)
Despite what you think, this is exactly the "barn door" paradox I mentioned above, and the apparent paradox is resolved by recognizing that the speeds and distances are such so that the events at opposite ends of the match (or ladder) are disassociated, and observers at each end are having their observations distorted by latency due to the speed of light.
posted by Steven C. Den Beste at 2:05 AM on December 7, 2006
So referencing the time scales of two inertial frames of reference requires them to be close enough together physically so that errors induced by latency due to speed-of-light are negligible within the context of the kinds of measurements that are being made.
I purposefully didn't give an absolute distance, because it really is a function of the measurements being made.
The two ends of the match are far enough apart, as a function of the kinds of measurements being made here, so that there's a non-negligible error in measurements between the two.
These movements are timed so that the match meets the matchbox at same time and place.
It isn't the "same time and place". It's two different places whose times can't be correlated to one another. Observers at each end of the match will observe somewhat different things. What each sees will be completely consistent, but they may disagree with one another.
(In extreme cases you can't even necessarily assume simultaneity over sub-atomic distances.)
Despite what you think, this is exactly the "barn door" paradox I mentioned above, and the apparent paradox is resolved by recognizing that the speeds and distances are such so that the events at opposite ends of the match (or ladder) are disassociated, and observers at each end are having their observations distorted by latency due to the speed of light.
posted by Steven C. Den Beste at 2:05 AM on December 7, 2006
Steven C. Den Beste: "Despite what you think, this is exactly the "barn door" paradox I mentioned above, and the apparent paradox is resolved by recognizing that the speeds and distances are such so that the events at opposite ends of the match (or ladder) are disassociated, and observers at each end are having their observations distorted by latency due to the speed of light."
Hmmm are you saying that the observer sitting in the middle of the match will see the match entering on an angle? I still don't believe this is quite the same as the barn example. In the barn example, an observer on the pole sees the barn doors close at different times. That is the resolution of the paradox in that case.
posted by zaebiz at 2:24 AM on December 7, 2006
Hmmm are you saying that the observer sitting in the middle of the match will see the match entering on an angle? I still don't believe this is quite the same as the barn example. In the barn example, an observer on the pole sees the barn doors close at different times. That is the resolution of the paradox in that case.
posted by zaebiz at 2:24 AM on December 7, 2006
See this page and look at the man falling into grate section. I always find it very helpful to remember that the objects in these things aren't single solid objects but are inherently bendy and compressible thanks to the limits put on them by SR.
posted by edd at 2:49 AM on December 7, 2006
posted by edd at 2:49 AM on December 7, 2006
edd has it right.
Ignoring time effects, there's nothing different about this scenario than with Galilean relativity. Take a box full of some chemicals and run with it. You're adding to the kinetic energy of the particles, so you might think the temperature is higher and the reaction should go faster.
It doesn't. For purposes of reactions, temperature is defined as the average kinetic energy of particles in the frame where the average momentum is zero, i.e. the "rest frame" of the reagents. Just like mass is defined as the mass in the rest frame.
Adding in time dilation is trivial once you know how to deal with all the other "paradoxes" of special relativity. The twin paradox is a perfect example. Aging is really just a string of chemical reactions, and it's easily dealt with. One twin comes back younger. That's fine. Now put one ice cube on a rocket at room temp, leave another home at room temp, and send the rocket off. When it comes back, you may find the ice that stayed home melted while the ice that went away didn't. No discussion of relativistic temperature effects is necessary.
posted by dsword at 8:04 AM on December 7, 2006
Ignoring time effects, there's nothing different about this scenario than with Galilean relativity. Take a box full of some chemicals and run with it. You're adding to the kinetic energy of the particles, so you might think the temperature is higher and the reaction should go faster.
It doesn't. For purposes of reactions, temperature is defined as the average kinetic energy of particles in the frame where the average momentum is zero, i.e. the "rest frame" of the reagents. Just like mass is defined as the mass in the rest frame.
Adding in time dilation is trivial once you know how to deal with all the other "paradoxes" of special relativity. The twin paradox is a perfect example. Aging is really just a string of chemical reactions, and it's easily dealt with. One twin comes back younger. That's fine. Now put one ice cube on a rocket at room temp, leave another home at room temp, and send the rocket off. When it comes back, you may find the ice that stayed home melted while the ice that went away didn't. No discussion of relativistic temperature effects is necessary.
posted by dsword at 8:04 AM on December 7, 2006
I really don't think it's quite that clear. [PDF]
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posted by kickingtheground at 8:39 AM on December 7, 2006
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posted by kickingtheground at 8:39 AM on December 7, 2006
Well, I don't know if "average velocity" of particles is a good definition for temperature. After all, you can have Bose-Einstein condensates (very-cold substances) whizzing around the sun at inconcievable velocities on a human scale. And the average velocity of boiling water should be low compared to the pot it sits in.
One way of looking at temperature is that it is a measurement of states of thermal equilibrium of a system relative to other systems. If a relativistic astronaut sucked on an oral thermometer, the temperature of both will adjust to an equilibrium point of 37C or 98.6F. Both are traveling at the same speed and have similar relativistic effects.
A second way of looking at temperature is that it is related to the entropy or disorder of a system. A collection of objects will be just as disordered at C-1 as at 0 assuming that all objects are being affected by the same forces. Some particle accelerators have cooling rings to reduce the disorder within a particle stream traveling at realitivistic speeds. The cooling ring eliminates particles that are going faster or slower than the ideal speed.
posted by KirkJobSluder at 9:27 AM on December 7, 2006
One way of looking at temperature is that it is a measurement of states of thermal equilibrium of a system relative to other systems. If a relativistic astronaut sucked on an oral thermometer, the temperature of both will adjust to an equilibrium point of 37C or 98.6F. Both are traveling at the same speed and have similar relativistic effects.
A second way of looking at temperature is that it is related to the entropy or disorder of a system. A collection of objects will be just as disordered at C-1 as at 0 assuming that all objects are being affected by the same forces. Some particle accelerators have cooling rings to reduce the disorder within a particle stream traveling at realitivistic speeds. The cooling ring eliminates particles that are going faster or slower than the ideal speed.
posted by KirkJobSluder at 9:27 AM on December 7, 2006
kickingtheground:
There seem to be two distinct questions. One involves what happens to the rates of chemical reactions, and the other concerns the distribution you measure.
I think the answer to the first is clear, and I don't see any challenge of it in the papers you link: in the rest frame of the beaker where some reaction occurs, things go along at a specific rate. In any other frame, that rate will be given by a simple Lorentz transformation. If you define temperature as being the parameter that gives you the right rate in the rest frame, then it's clear how temperature transforms: in whatever way gives you a simple time-dilated rate. Whether there's a good way to write this down depends on the mathematical form of the rate law.
The second question deals with how to transform the distribution of energy one sees from a black body of a given "temperature." I don't understand why anyone would expect that there'd be a simple transformation law for this definition of temperature (i.e. the value of the parameter in the Planck distribution that gives you the right energy spectrum), in fact, I'd be amazed if there were a simple transformation; I think the bottom of the third to the top of the fourth page of your last linked paper makes this pretty clear. The comments on the importance of the definition of temperature are especially noteworthy: it really comes down to semantics. From the standpoint of figuring what you would measure (i.e. the number of photons with energy E), though, there's at least one operational procedure which should give you the right answer: go to the rest frame, write down the Planck distribution, then boost individually over each element of phase space (note that you have to consider the directions of the photons). This seems to be essentially what they have done in the last paper.
posted by dsword at 10:01 AM on December 7, 2006
There seem to be two distinct questions. One involves what happens to the rates of chemical reactions, and the other concerns the distribution you measure.
I think the answer to the first is clear, and I don't see any challenge of it in the papers you link: in the rest frame of the beaker where some reaction occurs, things go along at a specific rate. In any other frame, that rate will be given by a simple Lorentz transformation. If you define temperature as being the parameter that gives you the right rate in the rest frame, then it's clear how temperature transforms: in whatever way gives you a simple time-dilated rate. Whether there's a good way to write this down depends on the mathematical form of the rate law.
The second question deals with how to transform the distribution of energy one sees from a black body of a given "temperature." I don't understand why anyone would expect that there'd be a simple transformation law for this definition of temperature (i.e. the value of the parameter in the Planck distribution that gives you the right energy spectrum), in fact, I'd be amazed if there were a simple transformation; I think the bottom of the third to the top of the fourth page of your last linked paper makes this pretty clear. The comments on the importance of the definition of temperature are especially noteworthy: it really comes down to semantics. From the standpoint of figuring what you would measure (i.e. the number of photons with energy E), though, there's at least one operational procedure which should give you the right answer: go to the rest frame, write down the Planck distribution, then boost individually over each element of phase space (note that you have to consider the directions of the photons). This seems to be essentially what they have done in the last paper.
posted by dsword at 10:01 AM on December 7, 2006
To add to the confusion:
There are multiple definitions of "temperature," which, it turns out in everyday situations, give basically equivalent answers. But this equivalence relationship breaks down under relativisitic conditions. Part of asking what "temperature" we on earth see on the spaceship is picking a definition of temperature - and it seems there is still debate on what one should expect to see even after picking a definition.
So really, the answer is, what "temperature" the Earthman measures on the ship depends, but yes, he'll see them having a lower r.m.s. velocity.
The key is that the temperature the rest observer measures doesn't matter - all that matters is what the spaceman measures. If the Earth observer peeks in the spaceship window when it passes, he may measure a different "temperature" if he's defining that by particle velocity but he'll also see other system parameters affected relativistically and this'll result in his go/no-go answer being the same as the spaceman's. Earthguy's "remote" thermometer will read a different number than Spaceguy's, but Spaceguy's is the number that matters.
posted by Opposite George at 3:08 PM on December 7, 2006
There are multiple definitions of "temperature," which, it turns out in everyday situations, give basically equivalent answers. But this equivalence relationship breaks down under relativisitic conditions. Part of asking what "temperature" we on earth see on the spaceship is picking a definition of temperature - and it seems there is still debate on what one should expect to see even after picking a definition.
So really, the answer is, what "temperature" the Earthman measures on the ship depends, but yes, he'll see them having a lower r.m.s. velocity.
The key is that the temperature the rest observer measures doesn't matter - all that matters is what the spaceman measures. If the Earth observer peeks in the spaceship window when it passes, he may measure a different "temperature" if he's defining that by particle velocity but he'll also see other system parameters affected relativistically and this'll result in his go/no-go answer being the same as the spaceman's. Earthguy's "remote" thermometer will read a different number than Spaceguy's, but Spaceguy's is the number that matters.
posted by Opposite George at 3:08 PM on December 7, 2006
So really, the answer is, what "temperature" the Earthman measures on the ship depends, but yes, he'll see them having a lower r.m.s. velocity.
i.e., depends on what he's calling temperature -- and b.t.w., in some cases it isn't obvious what he will measure.
posted by Opposite George at 3:11 PM on December 7, 2006
i.e., depends on what he's calling temperature -- and b.t.w., in some cases it isn't obvious what he will measure.
posted by Opposite George at 3:11 PM on December 7, 2006
and, duh, "he'll see them" -> "he'll see the molecules as"
posted by Opposite George at 3:15 PM on December 7, 2006
posted by Opposite George at 3:15 PM on December 7, 2006
Steven C Den Beste: I have to take issue with your handling of the twin paradox-- this statement in particular: "Nope! The one in the space ship had to accelerate in order to make a round trip, so it's him that suffers from time compression. It will unambiguously be the one in the space ship who is younger."
Change the statement of the problem slightly to remove acceleration: Two twins on Earth, twenty years old. One sets out at 0.6c. From the Earth frame of reference, the outbound twin travels for five years (for a distance of 3 lightyears). At this point, he has aged four years (time dilation factor 1.25) and is celebrating his 24th birthday when he gets a call from a girl who is nearby, heading the opposite direction at the same speed (again, Earth-frame), also celebrating her 24th birthday. They high-five en-route (tricky- artificial limbs required afterward). The girl returns to Earth on her 28th birthday, just as the homebound twin is celebrating his 30th.
That was all in the Earth-frame. Now, look at it in the outbound twin's frame:
He sits in place for 4 years while the Earth shoots away from him at 0.6 c, so it is 2.4 l.y. away from him. He thinks of his twin back home and calculates that he is only 23.2 years old because of time dilation. The 24 year old girl passes him at 15/17 c, and so spends the next 8.5 years catching up to the earth. Due to her speed, she ages 2.125 times slower than outbound twin, and so is 28 when she reaches earth. In the same period of time, the earthbound twin has aged 8.5/1.25=6.8 years, and so (again) is 30.
The same calculations can be done from the perspective of the girl.
Working through this problem from the different perspectives and finding that the results at the meeting points are the same in each case, despite the widely varying interpretations of distances, speeds and times, is what finally allowed me to feel that I had a solid grasp of special relativity.
posted by alexei at 3:36 PM on December 7, 2006
Change the statement of the problem slightly to remove acceleration: Two twins on Earth, twenty years old. One sets out at 0.6c. From the Earth frame of reference, the outbound twin travels for five years (for a distance of 3 lightyears). At this point, he has aged four years (time dilation factor 1.25) and is celebrating his 24th birthday when he gets a call from a girl who is nearby, heading the opposite direction at the same speed (again, Earth-frame), also celebrating her 24th birthday. They high-five en-route (tricky- artificial limbs required afterward). The girl returns to Earth on her 28th birthday, just as the homebound twin is celebrating his 30th.
That was all in the Earth-frame. Now, look at it in the outbound twin's frame:
He sits in place for 4 years while the Earth shoots away from him at 0.6 c, so it is 2.4 l.y. away from him. He thinks of his twin back home and calculates that he is only 23.2 years old because of time dilation. The 24 year old girl passes him at 15/17 c, and so spends the next 8.5 years catching up to the earth. Due to her speed, she ages 2.125 times slower than outbound twin, and so is 28 when she reaches earth. In the same period of time, the earthbound twin has aged 8.5/1.25=6.8 years, and so (again) is 30.
The same calculations can be done from the perspective of the girl.
Working through this problem from the different perspectives and finding that the results at the meeting points are the same in each case, despite the widely varying interpretations of distances, speeds and times, is what finally allowed me to feel that I had a solid grasp of special relativity.
posted by alexei at 3:36 PM on December 7, 2006
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[Not a physicist.]
posted by kickingtheground at 9:09 PM on December 6, 2006