Does the mass of an object increase in a gravitational field?

In summary: In this planet experiment you are comparing a gravitational acceleration where G is defined using time local to the planet to a centripetal acceleration caclulated using your watch. You know that your watch and the clocks on the planet (used to define G) run at different rates. But you are ignoring this and attributing the resultant mismatch in measurements to a change in mass. From the point of view of the observer in a strong gravitational field (near a black hole for example) according to her watch the Earth is rotating too fast. So fast in fact according to the force law it should be shedding it's oceans, unless it's relativistic mass had decreased proportionally to the amount of time dilation.
  • #1
Buckethead
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From a frame of reference outside of a gravitational field, does the mass of an object near a gravitational field increase?
 
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  • #2
No it doesn't. Mass is invariant which means it is not coordinate dependent.
 
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  • #3
timmdeeg said:
No it doesn't. Mass is invariant which means it is not coordinate dependent.
Just to be clear, I'm not referring to rest mass. Since time dilates and mass increases for an object that has a velocity relative to an observer, it lead me to wonder if mass also increases (since time dilates) for an object in a gravitational field relative to an observer outside that field.
 
  • #4
Buckethead said:
mass increases for an object that has a velocity relative to an observer,
Please don't say that. "Mass" means invariant mass. Say "relativistic mass" if you must use the term; better to call it total energy (divided by ##c^2## if you like).

How are you planning to measure the mass of this thing from far away?
 
  • #5
Ibix said:
Please don't say that. "Mass" means invariant mass. Say "relativistic mass" if you must use the term; better to call it total energy (divided by ##c^2## if you like).

How are you planning to measure the mass of this thing from far away?

Sorry, I will refer to it as relativistic mass then.

To measure the relativistic mass, I would measure the time of oscillation of a clock spring for example relative to my watch. Since the watch fulcrum would be moving more slowly, it would indicate it's mass would have to have increased so as not to violate m=f/a since the force available in the spring itself would be constant.
 
  • #6
Buckethead said:
Since time dilates and mass increases for an object that has a velocity relative to an observer, it lead me to wonder ...
The concept of relativistic mass is misleading, instead
The relativistic energy-momentum equation
is sufficient.
 
  • #7
So you are going to combine a time measurement in one place with a measurement of the spring constant (which includes a dimension of time) in another. That won't produce anything meaningful.
 
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  • #8
Ibix said:
So you are going to combine a time measurement in one place with a measurement of the spring constant (which includes a dimension of time) in another. That won't produce anything meaningful.

It seems meaningful in this regard: From the point of view of the observer in a strong gravitational field (near a black hole for example) according to her watch the Earth is rotating too fast. So fast in fact according to the force law it should be shedding it's oceans, unless it's relativistic mass had decreased proportionally to the amount of time dilation.

edit: Actually I meant it would have to increase its relativistic mass to hold the oceans in at such a high rotational velocity.
edit again: forget that last edit, it seems the mass must decrease. My point is, something has to give to prevent stresses from changing, either the mass or the force.
 
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  • #9
In this planet experiment you are comparing a gravitational acceleration where G is defined using time local to the planet to a centripetal acceleration caclulated using your watch. You know that your watch and the clocks on the planet (used to define G) run at different rates. But you are ignoring this and attributing the resultant mismatch in measurements to a change in mass. It's no more a change in mass than it would be if I calculated the centripetal acceleration in metres per minute squared, the gravitational acceleration in metres per second squared, and blamed the resulting factor of 3600 on a mass change.

So neither the mass nor the force has changed. You're just measuring the force wrongly because you are not accounting for the way your instruments work.
 
  • #10
Ibix said:
In this planet experiment you are comparing a gravitational acceleration where G is defined using time local to the planet to a centripetal acceleration caclulated using your watch. You know that your watch and the clocks on the planet (used to define G) run at different rates. But you are ignoring this and attributing the resultant mismatch in measurements to a change in mass.

Thank you for this clarification. Doesn't this indicate that if I correct this error and use my time to determine the value of G, then it would be different than the (planets) local value of G in which case it is F that changes along with t in the force equation when all measurements are performed using only my time and that it is this then that prevents the planet from ripping apart?
 
  • #11
Again, you are just using funny time units. They're consistent this time, so things will work out. The numbers are different, but they'd be different if you switched to using minutes instead of seconds.

It's not recommended to do this in relativity because the ratio of my clock rate to yours isn't generally well defined in curved spacetime. You get away with it here because it's possible to write the Schwarzschild metric in a time-independent way. In general, though, you need to make all measurements (time and otherwise) locally or derive the local measurements from your remote observations.
 
  • #12
Ibix said:
Again, you are just using funny time units. They're consistent this time, so things will work out. The numbers are different, but they'd be different if you switched to using minutes instead of seconds.

It's not recommended to do this in relativity because the ratio of my clock rate to yours isn't generally well defined in curved spacetime. You get away with it here because it's possible to write the Schwarzschild metric in a time-independent way. In general, though, you need to make all measurements (time and otherwise) locally or derive the local measurements from your remote observations.

I'm grateful for your help in getting this sorted out for me. The reason I needed to get this clear is because I'm giving a short philosophical talk on "The Flow of Time" and am trying to predict questions I may get. Central to my talk will be that time as we know it is the relationship between things that change and according to Leibniz (paraphrasing) "Time can only be defined as a sequence of events" and further argued "There is no such thing as an autonomous time" (Source: Carlo Rovelli in "The Order of Time").

Since time is only an order of events without an autonomous time it could be argued the the rate of change of the universe as a whole might be undefinable. The question I predict will come up might be along the lines of "If the universe as a whole changed too quickly, wouldn't things just fly apart?" I suspected that something in Newtons force law must change in order for things to remain stable hence my question about whether it might be the mass that changes. I was planning on using the example of an observer near a black whole as an example.
 
  • #13
If you have time before your lecture for research, Lee Smolin and James Gleick have written popular books on time that you could skim at the library. Please consider following the mentors' suggestions: mass particularly change in mass may not be the best concept to help you explain "Flow of Time".

Look again at post #4 from Ibix. Big hint: Energy may be a useful concept to help understand time, rather than mass. Thanks.
 
  • #14
Klystron said:
If you have time before your lecture for research, Lee Smolin and James Gleick have written popular books on time that you could skim at the library.

Thank you for suggesting that. I read "Time Reborn" about a year ago, and again about 2 months ago and probably need to read it a third time. One thing I was disappointed in was how he only fleetingly mentioned relativity of simultaneity in relation to his new theory. It seems they are at odds with each other and I was looking for a good resolution to that in the book. I admire Smolin very much. I think he is a maverick thinker. I also enjoyed his book "The Trouble With Physics" which I also read twice. With regard to Gleick, are you referring to his book "Time Travel"?

Klystron said:
Please consider following the mentors' suggestions: mass particularly change in mass may not be the best concept to help you explain "Flow of Time".

Look again at post #4 from Ibix. Big hint: Energy may be a useful concept to help understand time, rather than mass. Thanks.

Absolutely! Believe me, I always take seriously the mentor's suggestions. Sometimes I feel like I'm talking to Einstein himself when I ask questions in this forum.

I'm hoping I will not have to dive that deep in my talk, but with Ibex's post #9 as a guide ("In this planet experiment you are comparing a gravitational acceleration where G is defined using time local to the planet...") I can see that G, being defined in terms of time changes using the observer's (near the black hole) time. And this resolves the problem of why the planets don't disintegrate if they are spinning faster.

With regard to your suggestion: "Energy may be a useful concept to help understand time, rather than mass", could you elaborate a little? How can calling mass energy via E=mc^2 help with explaining the flow of time?

I'm excited about my talk but still have a lot of research to do.
 
  • #15
Buckethead said:
To measure the relativistic mass

The term "relativistic mass" is not correct for the (hypothetical, it doesn't actually happen--see below) effect you are describing a measurement of. "Relativistic mass" refers specifically to an effect of relative motion. The mass inside the gravitational field in your scenario is at rest relative to the observer well outside the field. So any effect on mass due to a gravitational field, such as you are asking about (and there actually isn't any--see below) would not be correctly described as "relativistic mass".

Buckethead said:
Since the watch fulcrum would be moving more slowly, it would indicate it's mass would have to have increased so as not to violate m=f/a since the force available in the spring itself would be constant.

This is not correct. Force "redshifts" the same way energy does, i.e., by the time dilation factor, so the slower motion is due to the reduced force, not to any change in mass.

Buckethead said:
From the point of view of the observer in a strong gravitational field (near a black hole for example) according to her watch the Earth is rotating too fast.

You are making the common mistake of throwing around different scenarios without first getting a proper understanding of the original one. Throwing in the black hole plus the rotating Earth is too much. You need to get a handle on the simplest case--a non-rotating source of gravity and a test mass in its gravitational field--first.
 
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  • #16
Buckethead said:
From a frame of reference outside of a gravitational field, does the mass of an object near a gravitational field increase?
Maybe there is such an higher order effect :
https://en.wikipedia.org/wiki/Frame-dragging
"Static mass increase is a third effect noted by Einstein in the same paper.[6] The effect is an increase in inertia of a body when other masses are placed nearby. While not strictly a frame dragging effect (the term frame dragging is not used by Einstein), it is demonstrated by Einstein that it derives from the same equation of general relativity. It is also a tiny effect that is difficult to confirm experimentally."
 
  • #17
PeterDonis said:
The term "relativistic mass" is not correct for the (hypothetical, it doesn't actually happen--see below) effect you are describing a measurement of. "Relativistic mass" refers specifically to an effect of relative motion. The mass inside the gravitational field in your scenario is at rest relative to the observer well outside the field. So any effect on mass due to a gravitational field, such as you are asking about (and there actually isn't any--see below) would not be correctly described as "relativistic mass".

OK, great. Thank you for clarifying that.

PeterDonis said:
This is not correct. Force "redshifts" the same way energy does, i.e., by the time dilation factor, so the slower motion is due to the reduced force, not to any change in mass.

Interesting. This would correlate with what Idex said in post #9 about G changing due to time dilation. So it does seem things "pony up" to keep Newtonian laws working in the way I had hoped they would.

PeterDonis said:
You are making the common mistake of throwing around different scenarios without first getting a proper understanding of the original one. Throwing in the black hole plus the rotating Earth is too much. You need to get a handle on the simplest case--a non-rotating source of gravity and a test mass in its gravitational field--first.

I'm sorry, but I don't understand this. Without the black hole plus the rotating Earth I have no question. I have to use the two together as that is my question. Just to be clear, my core question now (not my original question) is this: An observer in a gravitational field notes that the Earth (outside the strong gravitational field of the observer) is spinning too fast (or a watch is ticking too fast). This observation is a real event. i.e. Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches. That being the case, I find Newtonian laws would be violated unless something else also changed. If this were not the case, the Earth would fly apart from spinning too fast. I originally thought it might be Mass that changed, but this error has been corrected. It seems instead that it is the force that changes which includes the force in a watch spring and the force due to G and in any other force. The so called "redshift" in the force. Is this correct?
 
  • #18
Buckethead said:
This would correlate with what Idex said in post #9 about G changing due to time dilation.

G doesn't change due to time dilation. I had missed that post of @Ibix, I'll need to take a look at it and respond separately.

Buckethead said:
it does seem things "pony up" to keep Newtonian laws working in the way I had hoped they would.

Locally, yes.

Buckethead said:
An observer in a gravitational field notes that the Earth (outside the strong gravitational field of the observer) is spinning too fast (or a watch is ticking too fast).

What do you mean, "too fast"? How does the observer in the gravitational field know this?

Buckethead said:
Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches.

The difference in elapsed time on watches of people who followed different paths through spacetime does not mean "time is actually moving faster". It just means their paths through spacetime had different lengths. It's no different from two people taking two routes from New York to Los Angeles and finding different elapsed distances on their odometers when they meet again: that doesn't mean "distance is actually moving faster" for one of them.

Buckethead said:
That being the case, I find Newtonian laws would be violated unless something else also changed.

Nope. Locally the laws are the same everywhere. But you are trying to apply local laws to situations that aren't local. That doesn't work.

Buckethead said:
I originally thought it might be Mass that changed, but this error has been corrected. It seems instead that it is the force that changes which includes the force in a watch spring and the force due to G and in any other force.

Neither the mass nor the force changes locally. But you are trying to describe things that are happening not locally, but distant from you. And you can't expect such a description to follow all of the same rules as a local description.
 
  • #19
Ibix said:
G is defined using time local to the planet

G isn't defined using anyone's local notion of time. It's a universal constant. So the explanation you are offering, that the distant observer's observations should be attributed to a change in G, does not work.
 
  • #20
Buckethead said:
It seems instead that it is the force that changes which includes the force in a watch spring and the force due to G and in any other force. The so called "redshift" in the force. Is this correct?

Not for the question you are asking now. The answer to the question you are asking now is that the spin rate of the Earth that matters for determining the Earth's structure and whether it will hold together is the spin rate local to the Earth, not the apparent spin rate seen by a distant observer.

You appear to have a misconception about the laws of physics and the principle of relativity. The principle of relativity, in the context of GR, is that the laws of physics locally are the same. You are leaving out the "locally" and trying to plug numbers based on the observations of a distant observer into laws that apply locally. That doesn't work. (@Ibix made this same point to you back in post #7. You need to think about it carefully.)
 
  • #21
Buckethead said:
An observer in a gravitational field notes that the Earth (outside the strong gravitational field of the observer) is spinning too fast (or a watch is ticking too fast). This observation is a real event. i.e. Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches.
You have to be careful here - there are some some dubious assumptions hidden in the innocent-sounding words "real observation" and "too fast".

First, we need to be clear about exactly what we are observing: Suppose we position a strobe light on the surface of the earth, calibrated to flash once every revolution (according to an observer at rest relative to the center of the Earth and close enough to the Earth to also be considered "outside the string gravitational field". Now we, deep inside the gravitational field, will observe the time at which these flashes arrive. That is, our "real observation" is of the proper time along our worldline between the receipt of successive flashes, not the proper time the Earth takes to make one rotation.

We conclude from this observation (it helps if the transmitted light signals include the time on Earth at which they are sent) that the proper time the Earth takes to make one rotation is greater than the proper time between our receipt of successive flashes. But how does that have to mean that the Earth is spinning "too fast", or equivalently that too little time is passing between revolutions? We have two time intervals. One is longer than the other, but we can interpret this as the longer one being too long, the shorter one being too short, or (best) both of them being just right for what they are.

Note that if we're going to bring the clocks back together again the whole thing is just a variant of the twin paradox, with the same resolution: the Earth isn't spinning too fast, it's spinning at the same rate for a longer time.
 
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  • #22
Buckethead said:
The so called "redshift" in the force.

I should clarify what this means. Suppose you are sitting far away from a massive body like the Earth, and you want to measure the mass of an object deep inside the gravitational field of that body. You propose to measure it this way: assemble a mechanical linkage that will let you push on the mass from a distance by transmitting the force you exert to the mass. Then you measure the acceleration of the object by observing its motion and timing it with your own clock, and plug that and the force into ##m = F / a## to get the mass.

Since the object is time dilated with respect to you, the object's motion will appear slow, so ##a## will be smaller than what an observer next to the mass would measure. But the force ##F## that you have to apply will also be smaller by the same redshift factor, so you will still get the same ratio ##F / a## for the mass. The force will be smaller because, as it gets transmitted down the mechanical linkage, the linkage itself, which, heuristically, "wants" to fall in the gravitational field, will increase the force that finally gets applied locally to the mass. Or, to put it another way, an observer sitting next to the mass, at the bottom of the linkage, will measure a force that is larger, by the redshift factor, than the force you apply at the top of the linkage.

(Note that all of this is still just for this particular case; it is not a general argument that everything a distant observer observes gets adjusted by the same factor, nor is it a general argument that you can plug distant observations into the same laws that apply locally. You can't.)
 
  • #23
PeterDonis said:
G isn't defined using anyone's local notion of time. It's a universal constant. So the explanation you are offering, that the distant observer's observations should be attributed to a change in G, does not work.
I'm attributing it to a change of units in G, I think. If I am a distant observer using time on my wristwatch to analyse something happening at some more-or-less constant altitude far below me in a Schwarzschild metric, that's the same as an observer local to the experiment using a watch that ticks Schwarzschild coordinate time at that altitude, surely? It's just funny time units.

Edit: and you can only get away with it in stationary metrics, and as long as the experiment isn't changing r coordinate. And many other caveats, I'm sure.
 
  • #24
Ibix said:
I'm attributing it to a change of units in G, I think

No, it isn't.

Ibix said:
If I am a distant observer using time on my wristwatch to analyse something happening at some more-or-less constant altitude far below me in a Schwarzschild metric, that's the same as locally using a watch that ticks Schwarzschild coordinate time, surely?

No.

Ibix said:
It's just funny time units.

No, it's different curves through spacetime having different lengths. The units measuring the length along each curve are the same.

Consider the analogy I made in response to @Buckethead in post #18: two people travel from New York to Los Angeles along different routes. Does the fact that their odometers register different elapsed distances at the end mean they were using different units to measure distance?
 
  • #25
PeterDonis said:
No, it's different curves through spacetime having different lengths. The units measuring the length along each curve are the same.
I think that's a matter of interpretation of the original question. My reading of #5 is basically that Buckethead is looking down at a mass on a spring and using his own watch to time its oscillations. He isn't correcting for redshift, which is how he can say the spring is moving slowly.

I completely agree that you can (should!) use the same units to measure the interval between ticks of Buckethead's watch and cycles of the spring. But he isn't doing so - if he were doing then he'd be using the spring's proper time and would have no grounds to complain that the spring was oscillating at any unusual rate. You seemed to me to be making this point yourself in #18 when you asked What do you mean, "too fast"?
PeterDonis said:
Consider the analogy I made in response to @Buckethead in post #18: two people travel from New York to Los Angeles along different routes.
But I don't think that's a precise analogy to Buckethead's experiment. Buckethead's worldline and that of the spring never meet. It's more like Buckethead walking a line of constant latitude near the pole and wondering how he's staying ahead of a sprinter on the equator. What he really should be doing is using metre rules on the equator to measure the distance covered by the sprinter. But what he's doing is more like projecting the ends of his own metre rules along lines of constant longitude down to the equator. Which, in this highly symmetric case, would be a perfectly valid procedure if only he didn't call the projected length of his ruler a metre.
 
  • #26
Nugatory said:
That is, our "real observation" is of the proper time along our worldline between the receipt of successive flashes, not the proper time the Earth takes to make one rotation.

Just a quick clarification before I answer everyone's posts more extensively. Are you saying we will measure the pulses of light to occur at the rate of once per day proper time (i.e. according to the time on my watch)?
 
  • #27
Ibix said:
I think that's a matter of interpretation of the original question.

No, it isn't. In order for the redshift formula to be true, you have to use the same units to measure lengths along all curves. The OP simply hasn't thought through what he's asking.

Ibix said:
My reading of #5 is basically that Buckethead is looking down at a mass on a spring and using his own watch to time its oscillations.

Yes.

Ibix said:
He isn't correcting for redshift, which is how he can say the spring is moving slowly.

No. He is failing to think correctly about what "slowly" means. Slowly compared to what? Once that question is answered correctly, there is no room left for any change of units; the units have to be the same everywhere. (Note that your "correcting for redshift" implicitly answers the question "slowly compared to what"--if it didn't, the idea of "redshift" wouldn't make sense either. Redshift compared to what?)

Ibix said:
Buckethead's worldline and that of the spring never meet.

That's true, but in order to say anything about "redshift" or "watch ticking slowly" or anything else, you have to specify what events on the worldlines that never meet happen "at the same time"; in other words, you need to adopt a simultaneity convention. The one being used in this thread is the one corresponding to Schwarzschild coordinates (which actually can be given a coordinate-independent definition, but I don't think we need to go into that here). Once you have that, again, you are simply comparing lengths along different paths in spacetime and there is no room for any additional variability in units; you have to use the same units along all paths or all of the statements about "redshift", "ticking slowly", etc. don't work.
 
  • #28
Buckethead said:
Just a quick clarification before I answer everyone's posts more extensively. Are you saying we will measure the pulses of light to occur at the rate of once per day proper time (i.e. according to the time on my watch)?
No. Under the conditions you have specified, the reception events will be separated by less than one day of proper time along our worldline (assuming that a "day" is defined to be the number of cesium-clock seconds that will be counted by a clock on Earth during one rotation).

But there's not a lot of physical significance that you can attach to that fact; in particular there's no need to explain how the Earth can hold together even though it's turning "too fast". There are several ways of seeing this. One is to consider what would happen if we were to move up or down in the gravity well; the proper time between our reception events changes but obviously the Earth doesn't care. Another is to consider the twin paradox situation (which you introduced above): when we reunite the clocks we will find that the Earth clock shows N days elapsed while the Earth rotated N times so of course the Earth isn't rotating too fast.
 
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  • #29
Buckethead said:
I'm sorry, but I don't understand this. Without the black hole plus the rotating Earth I have no question. I have to use the two together as that is my question. Just to be clear, my core question now (not my original question) is this: An observer in a gravitational field notes that the Earth (outside the strong gravitational field of the observer) is spinning too fast (or a watch is ticking too fast). This observation is a real event. i.e. Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches. That being the case, I find Newtonian laws would be violated unless something else also changed. If this were not the case, the Earth would fly apart from spinning too fast. I originally thought it might be Mass that changed, but this error has been corrected. It seems instead that it is the force that changes which includes the force in a watch spring and the force due to G and in any other force. The so called "redshift" in the force. Is this correct?
Well how about an observer in a very deep gravity well observing a clock that is located far above, and wondering how the clock hands do not break, as they are going around like crazy?

We can see that extra mass can not be responsible of keeping the clock hands intact, right?

Then maybe we could consider a tug of war between two rockets, one next to that previous observer, and the other one next to the clock.

(We know the upper rocket is propelling reaction mass at crazy rate, I mean according to the observer)

(According to the observer the force pulling the rope upwards is 'crazy', to use that technical term again)

Here's something that might be interesting to @Buckethead:
If for some reason some pieces of the rope are connected together using gravitating masses, those masses need to be large at the observers end. Smaller masses are enough to keep the rope together at the other end where clocks run fast according to the observer.
 
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  • #30
Nugatory said:
No. Under the conditions you have specified, the reception events will be separated by less than one day of proper time along our worldline (assuming that a "day" is defined to be the number of cesium-clock seconds that will be counted by a clock on Earth during one rotation).

I'm confused. Isn't proper time my time (standing standing near this black hole)? If so why did you say "less than one day of proper time" only to say "assuming a day is a day as measured by an Earth bound clock? Sorry, I'm just not understanding. To be clear about my question. Using my watch standing near a black hole and timing the pulses as they come in from the distant Earth that is sending out the pulses once per Earth revolution, do I measure one day between pulses. I'm sorry to have to ask again, but I'm not getting something here.
 
  • #31
Buckethead said:
I'm confused. Isn't proper time my time (standing standing near this black hole)? If so why did you say "less than one day of proper time" only to say "assuming a day is a day as measured by an Earth bound clock?
Define a "day" to be the time measured by a cesium clock on Earth during one rotation of the Earth (using the definition of "rotation" I used in #21 above). That's some number of cesium clock cycles on earth. Your handy wristwatch cesium clock, deeper in the potential well, will count fewer cycles between consecutive pulses; that is, less than a day's worth of proper time will pass between the reception events.
 
  • #32
PeterDonis said:
What do you mean, "too fast"? How does the observer in the gravitational field know this?

By looking through a scope toward the Earth and seeing that it is spinning more than once a day according to the watch on the hand of the observer?
PeterDonis said:
The difference in elapsed time on watches of people who followed different paths through spacetime does not mean "time is actually moving faster". It just means their paths through spacetime had different lengths. It's no different from two people taking two routes from New York to Los Angeles and finding different elapsed distances on their odometers when they meet again: that doesn't mean "distance is actually moving faster" for one of them.

This is a good analogy. I suppose it's similar to saying the time traveler in H.G. Wells "The Time Machine" isn't really watching the candle in his lab burning down faster, he's only taking a different path through spacetime, a shortcut, to a point where the candle has burned down but taking less time to make the trip.

PeterDonis said:
Nope. Locally the laws are the same everywhere. But you are trying to apply local laws to situations that aren't local. That doesn't work.

I would have thought you could do a Lorentz transform from a non local environment to a local environment with the point of the exercise being, beside time contraction, what else would need to be transformed, giving you insight into what else changes beside time (mass or force). It seems you are suggesting such a transform is not possible
 
  • #33
Buckethead said:
By looking through a scope toward the Earth and seeing that it is spinning more than once a day according to the watch on the hand of the observer?

How does the observer know how long a "day" is? How does he know that the time 86,400 seconds has some special significance for the Earth? It doesn't have any special significance in his own observations.

Buckethead said:
I would have thought you could do a Lorentz transform from a non local environment to a local environment

You thought wrong. You can't. A Lorentz transformation only works between inertial frames, and in a curved spacetime (i.e., in the presence of gravity), there are no global inertial frames, only local ones. And there is no local inertial frame that covers both the observer and the Earth in your scenario. Indeed, there is no single local inertial frame that covers the Earth as a whole itself, even leaving out observers far away from it.
 
  • #34
Nugatory said:
We conclude from this observation (it helps if the transmitted light signals include the time on Earth at which they are sent) that the proper time the Earth takes to make one rotation is greater than the proper time between our receipt of successive flashes. But how does that have to mean that the Earth is spinning "too fast", or equivalently that too little time is passing between revolutions? We have two time intervals. One is longer than the other, but we can interpret this as the longer one being too long, the shorter one being too short, or (best) both of them being just right for what they are.

I see what you are saying now. Sorry for all the confusion, I get tripped up when I hear the term "proper time". So the pulses by my watch are coming in faster, which I would expect because of time dilation of the distant Earth, but that is not to say the Earth is moving faster through time than I am or that I am moving through time more slowly than Earth. With that in mind, it is invalid for me to say that the Earth is spinning to fast and there needs to be some compensation to prevent havok. Of course. That makes perfect sense. Actually it fits right in with the idea that the "flow of time" is an illusion and that only the relationships between changing events is what counts, the "sequence of events" as Leibniz puts it.

I have to digest all this to really get it, but I think I can see this all more clearly now.
 
  • #35
PeterDonis said:
How does the observer know how long a "day" is? How does he know that the time 86,400 seconds has some special significance for the Earth? It doesn't have any special significance in his own observations.

Yes, this is all making sense now. It's just a relationship between the two, not absolutes.
PeterDonis said:
You thought wrong. You can't. A Lorentz transformation only works between inertial frames, and in a curved spacetime (i.e., in the presence of gravity), there are no global inertial frames, only local ones

I didn't know this. So this is definitely not the same situation as doing Lorentz transforms between two passing ships. So two passing ships is considered a set of global inertial frames? But a frame in a gravitational field and one without are two frames that cannot share such a relationship? This is all very eye opening to me.
 
<h2>1. Does the mass of an object affect its gravitational pull?</h2><p>Yes, the mass of an object does affect its gravitational pull. The more mass an object has, the stronger its gravitational pull will be. This is because mass is one of the factors that determines the strength of the gravitational force between two objects.</p><h2>2. Does the mass of an object increase in a stronger gravitational field?</h2><p>No, the mass of an object does not increase in a stronger gravitational field. The mass of an object remains constant, regardless of the strength of the gravitational field it is in. However, the weight of an object may change in a stronger gravitational field, as weight is a measure of the force of gravity on an object.</p><h2>3. How does the mass of an object affect its acceleration in a gravitational field?</h2><p>The mass of an object does not affect its acceleration in a gravitational field. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that in a gravitational field, all objects will experience the same acceleration regardless of their mass.</p><h2>4. Does the mass of an object change when it is in orbit?</h2><p>No, the mass of an object does not change when it is in orbit. The mass of an object remains constant, regardless of its location or motion. However, the weight of an object may change in orbit due to the changing strength of the gravitational field.</p><h2>5. Can an object have a negative mass in a gravitational field?</h2><p>No, an object cannot have a negative mass in a gravitational field. Mass is a fundamental property of matter and it cannot be negative. Objects with negative mass would violate the laws of physics and are not possible in our universe.</p>

1. Does the mass of an object affect its gravitational pull?

Yes, the mass of an object does affect its gravitational pull. The more mass an object has, the stronger its gravitational pull will be. This is because mass is one of the factors that determines the strength of the gravitational force between two objects.

2. Does the mass of an object increase in a stronger gravitational field?

No, the mass of an object does not increase in a stronger gravitational field. The mass of an object remains constant, regardless of the strength of the gravitational field it is in. However, the weight of an object may change in a stronger gravitational field, as weight is a measure of the force of gravity on an object.

3. How does the mass of an object affect its acceleration in a gravitational field?

The mass of an object does not affect its acceleration in a gravitational field. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that in a gravitational field, all objects will experience the same acceleration regardless of their mass.

4. Does the mass of an object change when it is in orbit?

No, the mass of an object does not change when it is in orbit. The mass of an object remains constant, regardless of its location or motion. However, the weight of an object may change in orbit due to the changing strength of the gravitational field.

5. Can an object have a negative mass in a gravitational field?

No, an object cannot have a negative mass in a gravitational field. Mass is a fundamental property of matter and it cannot be negative. Objects with negative mass would violate the laws of physics and are not possible in our universe.

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