I Force due to acceleration and time flowing differently

[PeterDonis]

Which is why I wrote "forget about relativity for a while" or something along those lines

I'm not sure that's a good idea. The OP is asking about relativity, and the treatment of gravity in Newtonian mechanics vs. relativity is very different, so I'm not sure trying to start with a Newtonian treatment will be helpful.

If the scenario were modified to eliminate gravity (for example, taking place out in deep space far away from all gravitating bodies), then starting with a Newtonian treatment might work better.

 

[Dale]

The Lorentzian fundamental form doesn't induce a metric, because it's not positive definite

Physicists nevertheless call it a metric, even knowing that technically it is not

 

[malawi_glenn]

Physicists nevertheless call it a metric, even knowing that technically it is not

Not all of them ;)

 

[vanhees71]

Given the confusion of the OP, I think we should be a bit more pedantic. Our sloppy physicists' slang is not helpful!

 

[PeterDonis]

Given the confusion of the OP, I think we should be a bit more pedantic. Our sloppy physicists' slang is not helpful!

I'm not sure harping on the pedantic difference between a metric and a pseudometric is going to be the silver bullet that resolves the OP's confusion. IMO it's more likely to increase it.

 

[Dale]

Given the confusion of the OP, I think we should be a bit more pedantic. Our sloppy physicists' slang is not helpful!

Understood. That is where we differ. Given the confusion of the OP, I think the OP should study the basics some more and then come back to this question. I don't think being pedantic will resolve the confusion, but I do think studying the basics will.

 

[vanhees71]

That's of course true ;-). I still think that sloppy slang adds to the confusion. When I started to study relativity in high school this particular example of calling an indefinite quadratic form "a metric" took me quite some time to sort out again.

 

[HansH]

You don't accelerate frames. You accelerate objects. You can have a frame that is not an inertial frame, but talking about accelerating a frame doesn't make sense; accelerating a frame would mean the frame changes with time, but a frame already has to include time, so talking about it changing with time doesn't make sense.

Also, you can't .....

I see that I have to be much more accurate in my descriptions to prevent that the discussion drives away from the subject. as I understand An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed. so what I mean is then probably 2 rockets in space with an observer in each rocket going with relative speed =0 close to each other. Then they both synchronze their clocks and then one rocket accelerates for a certain amount of time with say 1g. so this gives a force of 1 Newton per kg mass at that mass during this amount of time. then after some time the same rocket accelerates for (I assume) a same amount of time with -1g so that after that there is no speed difference between the 2 rockets but only a certain distance in between. now they come together again by slow acelleration en de-aceleration and after that compare their watches. This means there is now a time difference between their watches related to the amount of time and acceleration that one of them acellerated. if I would use the double accellaration or the double time while acelerating finally the difference between their clocks would also be different. so therefore I suppose there could be an uderlying reason why both are related. (at least that both are related is clear and could be calculated)

 

[Vanadium 50]

Perhaps you should get a bit more comfortable with the basics of SR before moving on to accelerations?
Just a friendly tip.

yes of course, but I cannot prevent that the question pops up in my head at the wrong moment.

No, but you can't expect to start in the middle rather than the beginning and understand. That's why we call it the "beginning".

 

[HansH]

Understood. That is where we differ. Given the confusion of the OP, I think the OP should study the basics some more and then come back to this question. I don't think being pedantic will resolve the confusion, but I do think studying the basics will.

That sounds as a good compromise. For sure we differ as I am an engineer interested in the physics from pure interest relarted to topics I never studied so far (ok you have to make choices in your life) and most of you are professional physics guys. Only point for me is that I have a question about a possible relation that I see but no idea if or what underlying theory supports that, so what basics should I study then? looks like a chicken and egg. so there I need your advice. What I can do is study the links Dale gave and see if that makes sense, as I think that was a good advice.

 

[PeterDonis]

as I understand An inertial frame of reference is one in which the motion of a particle not subject to forces is in a straight line at constant speed

This definition is correct, but note that its actual meaning is different in relativity than in Newtonian physics. In Newtonian physics gravity is a force, but in relativity it is not. So in Newtonian physics a frame in which you, standing on the surface of the Earth, are at rest can be treated as an inertial frame, since the net force on you is zero (downward force of gravity exactly balances upward force of the Earth's surface pushing on you). But in relativity this is not the case; this frame is not inertial, because gravity is not a force so the only force on you is the Earth's surface pushing upward. In relativity, an inertial frame in your vicinity would be one in which a freely falling object, such as a rock dropping in vacuum, would be at rest.

one rocket accelerates for a certain amount of time with say 1g. so this gives a force of 1 Newton per kg mass at that mass during this amount of time. then after some time the same rocket accelerates for (I assume) a same amount of time with -1g

Time defined by what? By the rocket's clock? That's the simplest answer, but note that it is not the only possible answer, and other answers result in a different scenario being specified. (That would not be the case in Newtonian physics, which has absolute time, but there is no absolute time in relativity.)

after that there is no speed difference between the 2 rockets but only a certain distance in between. now they come together again by slow acelleration en de-aceleration and after that compare their watches. This means there is now a time difference between their watches related to the amount of time and acceleration that one of them acellerated

This is true, but it does not mean that acceleration "causes a difference in the rate of time flow". The fact that in this particular scenario, it happens to be possible to use the acceleration to compute the difference between the watches does not mean it is always possible to do that.

The only general rule that works is the geometric one: the difference between the watches is due to the difference in arc length along the two worldlines between the two meeting events. The particular scenario you describe just happens to be set up so that the difference in arc lengths, and hence the difference between the watches, has a definite relationship to the acceleration of one of the objects (and the time by that object's clock that the acceleration lasts). But that won't always be true.

 

[HansH]

No, but you can't expect to start in the middle rather than the beginning and understand. That's why we call it the "beginning".

The point is: if you are i a forest where you do not know the route, how do you then know where the beginning is? that only know the one that knows the route.

 

[PeroK]

The point is: if you are i a forest where you do not know the route, how do you then know where the beginning is? that only know the one that knows the route.

Most textbooks on SR start at the edge of the forest, not in the middle.

 

[Dale]

so what basics should I study then?

First, you should get comfortable with non-inertial frames in classical physics, and more importantly be comfortable using the Lagrangian for classical physics. Then you should study special relativity until you are comfortable with special relativity in inertial frames, and especially special relativity using the geometrical concept of four-vectors. It would also be helpful to have a passing familiarity with tensors.

With that background you will be prepared for understanding the nuances of the answers to this question.

if I would use the double accellaration or the double time while acelerating finally the difference between their clocks would also be different

I can produce scenarios where this is not true. You can have double the acceleration with the same difference between clocks.

 

[HansH]

The particular scenario you describe just happens to be set up so that the difference in arc lengths, and hence the difference between the watches, has a definite relationship to the acceleration of one of the objects (and the time by that object's clock that the acceleration lasts). But that won't always be true.

but then you agree that there is a relation. but as I understood so far we cannot explain that. probably the link of Dale can but I first need to check that (64 messages checking takes also time)

 

[PeterDonis]

but then you agree that there is a relation

There is a relation in this particular scenario. But that doesn't mean there is always such a relation in all scenarios. The question is whether you are just trying to understand this particular scenario, with the knowledge that what you learn about this particular scenario will not generalize to many other scenarios, or whether you want to learn the general rules that can be used in all scenarios.

 

[PeterDonis]

as I understood so far we cannot explain that

Depends on what you mean by "explain". Is there an "explanation" that will satisfy your current intuitions at your current level of knowledge? Quite possibly not. But that means you need to retrain your intuitions and increase your level of knowledge. It doesn't mean there is no explanation at all.

 

[HansH]

With that background you will be prepared for understanding the nuances of the answers to this question.

But as most people in this discussion have done these steps, I assume they should be able to draw a conclusion and translate that to a level that can be understood without first studying all the details. But I now understand that that is not possible or the answer is already given probably but too detailed for me?

 

[Dale]

I assume they should be able to draw a conclusion and translate that to a level that can be understood without first studying all the details

Your assumption is incorrect in this case, as evidenced by the disagreement on un-nuanced responses.

I know you want the shortcut, but I don't think there is one here. If I understand your question correctly then the second paper I provided answers it in what I think is the most basic way possible.

 

[PeterDonis]

translate that to a level that can be understood without first studying all the details

I gave you such a translation in post #61:

the difference between the watches is due to the difference in arc length along the two worldlines between the two meeting events.

"Difference in arc length" should be a geometrical concept that you can grasp.

 

[PeroK]

But I now understand that that is not possible or the answer is already given probably but too detailed for me?

It's the opposite. It's simpler. Proper time is the length of the path through spacetime. You embellish this with concepts such as "time slows down because of velocity/acceleration/force", which just confuses the issue.

 

[HansH]

I can produce scenarios where this is not true. You can have double the acceleration with the same difference between clocks.

I talked about acceleration during a certain time interval. so not ony the acceleration but also the duration of the accelleration. so it is about the combination of acceleration during a certain time interval (causing a speed difference between the 2 rockets) and the building up rate of the difference between the clocks. of course the difference in clocks builds up over the time that the speed difference remains, so the interval of keeping th speed difference before going back to same speed for both rockets also is in the relation. So I would expect you could write this down in a total equation something like :

difference in clocks= k1 x F1(build up speed difference) x F2(duration that the speed difference remans)

difference in clocks= k1 x F3(accelaration) x F4(duration the acceleration) x F2(duration that the speed difference remans)
so the question is: can we define such a relation where k1 is a constant and if so why does k1 then have the value as is has?

 

[Nugatory]

I cannot prevent that the question pops up in my head at the wrong moment. So I hoped for a clear high level answer already from people who are already comfortable with the matter.

However there is no answer that you will recognize as clear because, as this thread demonstrates, you lack the necessary background to follow a high-level answer.

There is also a problem with jumping into any question that just “pops up in [your] head”: as I said in your other thread

That’s a haphazard and disjointed activity that is unlikely to ever lead to a coherent understanding.
Taylor and Wheeler’s “Spacetime Physics” is available free online. Start at the first page. We can help you over the hard spots.

….and well before you get to the end, you’ll be able to handle this sort of question.

 

[HansH]

My preferred non-inertial reference frame is the radar coordinates which were popularized by Dolby and Gull. One thing that is nice about radar coordinates is that there is a known procedure for determining the metric given only the proper acceleration of the non-inertial observer. This procedure is a concrete answer to your question about the link between acceleration and the metric (which includes the "flow of time")

I have linked to two simple papers on this topic. I would recommend that you read both of them. If you cannot understand these papers then you are probably not yet ready for this question and should focus on the basics first.

This is indeed a no go for me. it already starts with fig 2 where it is not indicated what is on the axis and a lot of terms I am not familiar with : hypersurfaces of simultaneity, x µ = x µ (γ) (τ) etc.

 

[jbriggs444]

difference in clocks= k1 x F1(build up speed difference) x F2(duration that the speed difference remans)

Yes, the Lorentz transform contains a formula somewhat like that. $$t'=\gamma(t-\frac{vx}{c^2})$$
Your "duration the speed difference remains" is encoded in $x$"
Your "build up speed difference" is encoded in $v$ and $\gamma$.

The take away I get from this formula is that "There is a systematic synchronization skew based on displacement in the direction of relative travel. The higher the relative speed, the greater the skew".

Others take away a pithier form: "leading clocks lag".

 

[HansH]

Yes, the Lorentz transform contains a formula somewhat like that.

yes thanks, that helps. it is half of the discussion about the part including speed difference during a certain time. The other part should then be connected to the effects of building up this speed difference by an acceleration during a certain interval to build up the speed difference. and that results in the force in the other part of the equation I showed. So based on that both parts could be calculated so then it should be possible (at least for a physicist) to write down the complete equation. not sure if the part about building up the gamma factor during the acceleration phase. and then the k1 factor should follow.

 

[PeterDonis]

can we define such a relation

Only for the particular kind of scenario you are describing. It won't generalize to other scenarios. So, as I've already pointed out in post #66, the question is whether you want to learn a particular set of rules that will work in this particular scenario, but won't generalize, or whether you want to learn the general rules that will work for all scenarios. Which is it?

 

[PeroK]

yes thanks, that helps. it is half of the discussion about the part including speed difference during a certain time. The other part should then be connected to the effects of building up this speed difference by an acceleration during a certain interval to build up the speed difference. and that results in the force in the other part of the equation I showed. So based on that both parts could be calculated so then it should be possible (at least for a physicist) to write down the complete equation. not sure if the part about building up the gamma factor during the acceleration phase. and then the k1 factor should follow.

This is all wrong thinking - as has already been pointed out. Ultimately, differential ageing depends only on the velocity profiles in any IRF. Yes, velocity profiles can be deduced from proper acceleration profiles and hence from force profiles, but that's hardly the point. It's like trying to work out whether someone is speeding by looking at fuel consumption rather than the speedometer.

You are, of course, free to plough your own furrow. You are not the first and no doubt you won't be the last; but, it won't lead to an understanding of SR.

 

[PeterDonis]

Ultimately, differential ageing depends only on the velocity profiles in any IRF.

I would put this a bit differently: ultimately, differential aging depends only on the arc length along worldlines. In a situation where an inertial frame can be defined that covers the entire scenario, the arc length along worldlines can be computed as a function of the velocity profiles of the worldlines in that frame. The acceleration does not appear anywhere in this computation.

If one only knows the acceleration, one can of course compute the velocity profiles from the acceleration profiles, but that requirement would be specific to that particular scenario.

Also, not all scenarios in relativity have the property that an inertial frame can be defined that covers the entire scenario; all scenarios in special relativity do (since in SR there are always global inertial frames), but many scenarios in general relativity do not.

 

[HansH]

I can produce scenarios where this is not true. You can have double the acceleration with the same difference between clocks.

I talked about acceleration during a certain time interval. so not ony the acceleration but also the duration of the accelleration. so it is about the combination of acceleration during a certain time interval (causing a speed difference between the 2 rockets) and the building up rate of the difference between the clocks. of course the difference in clocks builds up over the time that the speed difference remains, so the interval of keeping th speed difference before going back to same speed for both rockets also is in the relation. So I would expect you could write this down in a total equation something like :

difference in clocks= k1 x F1(build up speed difference) x F2(duration that the speed difference remans)

difference in clocks= k1 x F3(accelaration) x F4(duration the acceleration) x F2(duration that the speed difference remans)
so the question is: can we

Only for the particular kind of scenario you are describing. It won't generalize to other scenarios. So, as I've already pointed out in post #66, the question is whether you want to learn a particular set of rules that will work in this particular scenario, but won't generalize, or whether you want to learn the general rules that will work for all scenarios. Which is it?

For understanding the general rules you need a 5 years physics study I suppose? That is at least not realistic in combination with my current situation in life. so yes good to learn the general rules, but probably not reachable. That I should have done then 30 years ago but would have given other compromises then also. But that is at least one thing I know from relativity that you cannot go back in time to make other choices. I was simply interested in the answer if there could be such a relation as I suspected, but that that turned out to be that difficult and probably not exsisting was not what I suspected.

 

[PeterDonis]

For understanding the general rules you need a 5 years physics study I suppose?

Go read post #70, where I stated a simple general rule that should not require 5 years of physics study to grasp, just a basic grasp of geometry.

 

[HansH]

post 70 refers back to post where you conclude: '
'the difference between the watches is due to the difference in arc length along the two worldlines between the two meeting events"
when I look what an arc lengt is that is the length of a trajectory in 4D spacetime as I understand but what do you mean by 2 meeting events? Isuppose you mean the trajectory of 2 observers that had different accelerations. but what is then a meeting event? and then the message you want to bring over from that senstence is still not clear to me in relation to different times on watches and force due to acceleration.
so what I now know is not sufficient to get the message from your general rule I suppose.

 

[anuttarasammyak]

ok but you can calculate the potential energy from the force working over a distance, so that makes the question not basically different I suppose.

Let us see the simple case of rotation. In the IFR where the axis of rotation is at rest, as SR is applicable, a rotating body undertakes time dilation of ratio
\gamma^{-1}=\sqrt{1-\frac{v^2}{c^2}}=\sqrt{1-\frac{r^2\omega^2}{c^2}}
where r is radius or rotation and $\omega$ is angular velocity.
In the non IFR system of the rotating body this ratio is interpreted as
\sqrt{g_{00}}=\sqrt{1+\frac{2\phi}{c^2}}
where
\phi=-\frac{1}{2}r^2\omega^2
, potential energy of centrifugal force per unit mass. A crew of the rotating whose system is non IFR observes that his lower potential energy than the center of rotation makes his time dilation.

 

[PeterDonis]

when I look what an arc lengt is that is the length of a trajectory in 4D spacetime as I understand

Yes.

but what do you mean by 2 meeting events?

The two events (points in spacetime) where the two observers meet and compare their watches. At the first event they ensure that their watches are synchronized; at the second event they compare the elapsed times on their watches and (in your scenario) find that they are different. The worldlines of the two observers must intersect that both of these points, so the difference in their elapsed times is just the difference in the arc lengths of their worldlines between the two points.

 

[HansH]

Yes.The two events (points in spacetime) where the two observers meet and compare their watches. At the first event they ensure that their watches are synchronized; at the second event they compare the elapsed times on their watches and (in your scenario) find that they are different. The worldlines of the two observers must intersect that both of these points, so the difference in their elapsed times is just the difference in the arc lengths of their worldlines between the two points.

ok clear. that is an easy to understand conclusion. (although I don't understand why but that is probably where the 5 years is for) but then given that difference in arc length, the next part of the story is about the question what happens during the sequence where both observers follow their trajectory while a certain acelleration as function of the position on each trajectory holds in order to make it possible to follow that trajectory and then the question if we can say something about that in order to determine a relation between that acelleration and the build up of the difference in stopwatches time. That is the most general way of looking to that problem I suppose?

 

[PeterDonis]

I don't understand why

Meaning you don't understand why the arc lengths are different? That's easy: because you are taking two different paths through spacetime between the same points. It's just geometry: it's the same as if you had two cars that took two different routes between, say, New York and Los Angeles. You would expect the elapsed odometer readings on the cars to be different because the routes were different. This is just the same thing in spacetime.

the next part of the story is about the question what happens during the sequence where both observers follow their trajectory while a certain acelleration as function of the position on each trajectory holds in order to make it possible to follow that trajectory

An object's trajectory through spacetime is determined by the geometry of spacetime and the object's 4-velocity. Proper acceleration is only relevant because it changes the object's 4-velocity.

in order to determine a relation between that acelleration and the build up of the difference in stopwatches time.

That relation will be different for different scenarios. In fact there are even scenarios in GR (in curved spacetime) where two objects that both have zero proper acceleration for all time have different elapsed times between two meetings.

That is the most general way of looking to that problem I suppose?

No. The most general way of looking at things is to look at spacetime geometry and object trajectories, and not to even bother about trying to find scenario-specific relationships between acceleration and other things, or even between velocity and other things.

Again, consider the analogy to two cars that take two different routes between New York and Los Angeles. The obvious general way of looking at such scenarios is to look at the geometry: the distance along each route. You could try to concoct relationships between that and, say, the compass direction in which each car points as a function of time along its route, or the rate of change of that compass direction, but any such relationship would be specific to the scenario and wouldn't really tell you anything useful.

 

[HansH]

Meaning you don't understand why the arc lengths are different?

no that is clear. what is unclear is how I can understand how that relates to difference between the watches at the meeting point and specifically how much time difference compared to arc length difference..

 

[PeterDonis]

is unclear is how I can understand how that relates to difference between the watches

The arc length difference is the difference between the watches. They're the same thing. Clocks record arc length along their worldlines in spacetime, just as odometers record arc length along their paths in ordinary space.

 

[HansH]

In fact there are even scenarios in GR (in curved spacetime) where two objects that both have zero proper acceleration for all time have different elapsed times between two meetings.

What do you mean by 'proper accelleration' When i am in freefall does that mean my proper accelleration is zero? so what for example about falling in freefall from the top of a building to earth. what is then my proper acceleration? is it then 0 or 1g?

 

[Dale]

This is indeed a no go for me. it already starts with fig 2 where it is not indicated what is on the axis and a lot of terms I am not familiar with : hypersurfaces of simultaneity, x µ = x µ (γ) (τ) etc.

Then I would recommend tabling this question for now and focusing on the basics. Do you understand Lagrangian mechanics? That will be generally helpful in learning physics.

Then the next thing to focus on is basic special relativity, especially four-vectors and the geometric interpretation of special relativity.

 

[PeroK]

Then the next thing to focus on is basic special relativity ...

... like the first postulate?

 

[HansH]

Do you understand Lagrangian mechanics?

I don't think that was part of my study somewhere.

 

[PeterDonis]

What do you mean by 'proper accelleration'

The acceleration measured by an accelerometer attached to the object.

When i am in freefall does that mean my proper accelleration is zero?

Yes.

 

[Dale]

I don't think that was part of my study somewhere.

If you want something useful, nothing you can study in relativity will be more useful than learning Lagrangian mechanics. And when you do learn the Lagrangian approach it will eventually make studying non-inertial reference frames easier. However, it isn't as "exciting" a topic as relativity.

 

[HansH]

Yes.

ok that helps. Then it means that I am not referring to the tem ''proper acceleration' but to the accelleration as result of an applied force such as a rocket engine. But then I think you are going to say that an observer cannot discriminate between gravity in curved spacetime and a rochet engine in flat spacetime due to the principle of equivalence. But at least for the acceleration he could isolate the effect of a rocket when he knows he has his rocket on or off as this gives the same relation between acceleration mass and force. so you could probably calculate the worldline with and without external forces and calculate only the contribution due to the external force to the difference in time between the watches. at least that should be easy for situations with flat spacetime as there ony the rocket could cause a change in speed.

 

[HansH]

If you want something useful, nothing you can study in relativity will be more useful than learning Lagrangian mechanics. And when you do learn the Lagrangian approach it will eventually make studying non-inertial reference frames easier. However, it isn't as "exciting" a topic as relativity.

thanks I will keep in mind.

 

[Dale]

Then it means that I am not referring to the tem ''proper acceleration' but to the accelleration as result of an applied force such as a rocket engine.

That is proper acceleration. An accelerometer does detect the acceleration as a result of a rocket engine. It seems like you may have a Newtonian physics misunderstanding here.

 

[HansH]

Yes. that would probably mean that the main idea of the topic cannot be united with general relativity due to the equivalence principle that cannot isolate an external acceleration from the effect of curvature on the path of a worldline? also based on the earlier statement that 2 different worldlines can exist with proper acceleration=0 over the whole worldline.

 

[Dale]

Yes. that would probably mean that the main idea of the topic cannot be united with general relativity due to the equivalence principle that cannot isolate an external acceleration from the effect of curvature on the path of a worldline?

I understand all of the words you are using but not the way you are using them.

Proper acceleration is the acceleration measured by an accelerometer. Both GR and Newtonian physics can describe the acceleration measured by an accelerometer. The difference between GR and Newtonian physics is not regarding proper acceleration. It is that Newtonian physics considers gravity to be a real force that is just coincidentally undetectable by accelerometers while GR considers gravity to be locally a fictitious force that is undetectable just like all other fictitious forces.

 

[HansH]

I understand all of the words you are using but not the way you are using them.

Proper acceleration is the acceleration measured by an accelerometer. Both GR and Newtonian physics can describe the acceleration measured by an accelerometer. The difference between GR and Newtonian physics is not regarding proper acceleration. It is that Newtonian physics considers gravity to be a real force that is just coincidentally undetectable by accelerometers while GR considers gravity to be locally a fictitious force that is undetectable just like all other fictitious forces.

I assume I am not talking about the difference between GR and Newtonian physics but the difference between GR with curved spacetime and special relativity with only flat spacetime. probably it helps to draw the worldlines in special relativity with the 2 rockets I used earlier and then calculate the difference between the stopwatches and compare that with the summed up product (integral) of acceleration times the time that the accelaration took at a certain value so integral(a(t)dt) (or probably a more relativistic equivalent that I cannot produce now) and divide that integral by the calculated timedifference between the stopwatches and see if this is a constant or not. But I agree that this should also work in general and for GR because otherwise it is not a valid theory. But proving that a theory is wong only needs one example.