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?
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.No it doesn't. Mass is invariant which means it is not coordinate dependent.
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).mass increases for an object that has a velocity relative to an observer,
Sorry, I will refer to it as relativistic mass then.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?
The concept of relativistic mass is misleading, insteadSince time dilates and mass increases for an object that has a velocity relative to an observer, it lead me to wonder ...
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.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.
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?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.
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").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.
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"?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.
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.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.
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".To measure the relativistic mass
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.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.
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.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.
Maybe there is such an higher order effect :From a frame of reference outside of a gravitational field, does the mass of an object near a gravitational field increase?
OK, great. Thank you for clarifying that.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".
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.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.
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?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.
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.This would correlate with what Idex said in post #9 about G changing due to time dilation.
Locally, yes.it does seem things "pony up" to keep Newtonian laws working in the way I had hoped they would.
What do you mean, "too fast"? How does the observer in the gravitational field know 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).
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.Time is actually moving faster and is a physical change that can be verified upon meeting up and comparing watches.
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.That being the case, I find Newtonian laws would be violated unless something else also changed.
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.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.
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.G is defined using time local to the planet
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.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?
You have to be careful here - there are some some dubious assumptions hidden in the innocent-sounding words "real observation" and "too fast".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.
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.The so called "redshift" in the force.
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.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.
No, it isn't.I'm attributing it to a change of units in G, I think
No.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, it's different curves through spacetime having different lengths. The units measuring the length along each curve are the same.It's just funny time units.
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.No, it's different curves through spacetime having different lengths. The units measuring the length along each curve are the same.
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.Consider the analogy I made in response to @Buckethead in post #18: two people travel from New York to Los Angeles along different routes.