Can different paths in spacetime have the same separation?

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  • #51
Don't forget that per your post 37 the components of the metric are functions of t. So any of them may be positive at T1 and negative at T2. This is, in fact, required for what you were considering through your barrage of posts last night.
 
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  • #52
If we consider the speed of a particle/point along our path [in consideration]it becomes infinitely large--both the coordinate speed and the physical speed. This is in view of the fact dt=0. This is valid even if the coefficients g(ii) take on unusual negative values.How does the path remain timelike in such a situation?

We may consider a limiting process speed -->an infinitely large quantity as dt--->0

One may avoid the problem by assuming [rather by imposing the condition]
{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2} is not positive.[at least for a time like path with dt=0 in the background]

If {-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2} is positive for a timelike path with dt=0 both timelike and spacelike paths will allow faster than light signals!
 
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  • #53
Anamitra said:
If we consider the speed of a particle/point along our path [in consideration]it becomes infinitely large--both the coordinate speed and the physical speed. This is in view of the fact dt=0. This is valid even if the coefficients g(ii) take on unusual negative values.How does the path remain timelike in such a situation?

We may consider a limiting process speed -->an infinitely large quantity as dt--->0

One may avoid the problem by assuming [rather by imposing the condition]
{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2} is not positive.[at least for a time like path with dt=0 in the background]

If {-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2} is positive for a timelike path with dt=0 both timelike and spacelike paths will allow faster than light signals!

The 4 velocity along a time like path is differentiation with respect to proper time (tau), and is always a unit vector. There is no such thing as velocity along a spacelike path. Differentiation with respect to coordinate t along a timelike path with constant value t is meaningless and undefined.
 
  • #54
PAllen said:
The 4 velocity along a time like path is differentiation with respect to proper time (tau), and is always a unit vector. There is no such thing as velocity along a spacelike path. Differentiation with respect to coordinate t along a timelike path with constant value t is meaningless and undefined.
Four velocity along a time like path is well defined and it conforms to a standard definition.How do you calculate it for the special timelike path under consideration for which dt=0? I mean to sayhow do you get the proper time interval to carry out the differentiation?

Possibly you mean to establish the idea that we may have exceptions for timelike paths for which the four velocity cannot be calculated.
 
  • #55
Anamitra said:
How do you calculate it for the special timelike path under consideration for which dt=0?
Why do you say this path is "timelike"? In most coordinate systems typically used in physics (Minkowski coordinates, for example), if dt=0 along a path then the path is spacelike, not timelike.
 
  • #56
We are analyzing a particular type of Time Like curve DaleSpam has tried to illustrate in #48[Path B]
Subsequent posts are concerned with such paths--Jesse should consider such special type of timelike paths in order to maintain the relevance of the discussion.
[You are requested to go through the subsequent posts[#48 and the following ones]
 
  • #57
Anamitra said:
Four velocity along a time like path is well defined and it conforms to a standard definition.How do you calculate it for the special timelike path under consideration for which dt=0? I mean to sayhow do you get the proper time interval to carry out the differentiation?

Possibly you mean to establish the idea that we may have exceptions for timelike paths for which the four velocity cannot be calculated.
Nothing special at all. d tau is based on the metric as you've specified; compute dt / d tau, dx / d tau, etc. All perfectly well defined for curve where d tau is positive (as it is along the curve under discussion).
 
  • #58
JesseM said:
Why do you say this path is "timelike"? In most coordinate systems typically used in physics (Minkowski coordinates, for example), if dt=0 along a path then the path is spacelike, not timelike.

This discussion started out with Anamitra proposing that the metric components could be functions including coordinate time, so that a 'similar' coordinate path could be spacelike in one region and timelike in another. Dalespam and I simplified to the case where you could have dt=0 along a path, yet it could, indeed, be spacelike in one region and timelike in another. The only thing 'special' about this situation is that it means that which coordinate(s) have time like character is different in different regions of spacetime. Perhaps if you read over the whole thread (mostly ignoring one of my misunderstandings, which Dalespam clarified), focusing on what Dalespam has said, you could clarify for Anamitra better than I've done so far.
 
  • #59
PAllen said:
Nothing special at all. d tau is based on the metric as you've specified; compute dt / d tau, dx / d tau, etc. All perfectly well defined for curve where d tau is positive (as it is along the curve under discussion).

The time component of the four velocity[As suggested by PAllen] is zero. Further differentiation wrt to propertime[ds] yields time component of the momentum vector[multiplication by rest mass is required] which is again zero. It is zero energy particle![Assuming a particle is capable of moving along a timelike curve/path]
 
  • #60
Anamitra said:
We are analyzing a particular type of Time Like curve DaleSpam has tried to illustrate in #48[Path B]
Subsequent posts are concerned with such paths--Jesse should consider such special type of timelike paths in order to maintain the relevance of the discussion.
[You are requested to go through the subsequent posts[#48 and the following ones]
Well, DaleSpam didn't specify any particular metric, but if you had one, then you could use it to compute proper time along the timelike path in the standard way, integrating \sqrt{g_{tt} dt^2 + g_{xx} dx^2 + g_{yy} dy^2 + g_{zz} dz^2 } along the path. In Dalespam's example only the x-coordinate varies so dt=dy=dz=0 along the path, meaning if the path varies from X1 to X2 and the path is timelike, you can calculate the proper time using the integral \int_{X1}^{X2} \sqrt{g_{xx} } \, dx. Does this answer your question "I mean to sayhow do you get the proper time interval to carry out the differentiation?"
 
  • #61
Anamitra said:
If we consider the speed of a particle/point along our path [in consideration]it becomes infinitely large
If a path is spacelike then it cannot represent the worldline of a particle. If a path is timelike then the velocity wrt any other (local) timelike path will be < c.
 
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  • #62
PAllen said:
Nothing special at all. d tau is based on the metric as you've specified; compute dt / d tau, dx / d tau, etc. All perfectly well defined for curve where d tau is positive (as it is along the curve under discussion).

For the Special Time Like path ,refered to by PAllen in the qouted section,the coordinate interval [time]between a pair of events is zero[since dt=0 for each and every subsection of the path].The proper time for such a path is the length of the path itself[ds=proper time interval]. If a person moves along this path with a clock in his hand what is he going to observe, considering the fact that dt=0 for each and every subsection of the path?
[A person should be capable of moving along a timelike path. Incidentally PAllen and DaleSpam are claiming the existence of a TIMELIKE PATH for which dt=0 for each infinitesimal subsection [ds^2>0 according to my convention].]
 
  • #63
Anamitra said:
For the Special Time Like path the coordinate interval [time]between a pair of events is zero[since dt=0 for each and every subsection of the path].The proper time for such a path is the length of the path itself[ds=proper time interval]. If a person moves along this path with a clock in his hand what is he going to observe, considering the fact that dt=0 for each and every subsection of the path?
Clocks measure proper time, not coordinate time. Different points on the path have different values of proper time, so he will see his clock tick forward as he moves along the path. In some coordinate system the worldline you are moving along right now has a constant t-coordinate, do you notice anything strange happening?
 
  • #64
JesseM said:
Clocks measure proper time, not coordinate time. Different points on the path have different values of proper time, so he will see his clock tick forward as he moves along the path. In some coordinate system the worldline you are moving along right now has a constant t-coordinate, do you notice anything strange happening?
Just tell me the value of propertime for the aforesaid path [between a pair of events on it]?Is it going to conform to the physical notion of the time interval as the person travels along the path between the events with his own clock?
[You should be careful enough to give due consideration to the nature of the very special type of the timelike of path we are dealing with--dt=0 for each and every infinitesimal subsection]
[The existence of such a path has been suggested by PAllen and DaleSpam]
 
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  • #65
Anamitra said:
Just tell me the value of propertime for the aforesaid path [between a pair of events on it]?
I did that. Did you read post #60?
Anamitra said:
Is it going to conform to the physical notion of the time interval as the person travels along the path between the events with his own clock?
Yes, of course. That's the physical meaning of "proper time"--it always corresponds to clock time, and "clock time"/"proper time" are both coordinate-independent notions (of course for a given spacetime geometry, the equations for the components of the metric change depending on what coordinate system you use in that spacetime, but when you integrate \sqrt{g_{tt} dt^2 + g_{xx} dx^2 + g_{yy} dy^2 + g_{zz} dz^2 } along a path using the correct form of the metric tailored to the coordinate system which you're using to describe the path, you will get the same answer regardless of what coordinate system you choose).
Anamitra said:
[You should be careful enough to give due consideration to the nature of the very special type of the timelike of path we are dealing with--dt=0 for each and every infinitesimal subsection]
[The existence of such a path has been suggested by PAllen and DaleSpam]
Coordinate time is completely irrelevant, only proper time matters if you want to know what physical clocks are doing. If you think it somehow makes a difference that dt=0, I think you're misunderstanding something basic about the physical meaning of proper time (and the fact that it's independent of what coordinate system you choose--we could easily describe the exact same physical path in a different coordinate system where dt was not zero, and the proper time along the path would necessarily be exactly the same).
 
  • #66
JesseM said:
Well, DaleSpam didn't specify any particular metric, but if you had one, then you could use it to compute proper time along the timelike path in the standard way, integrating \sqrt{g_{tt} dt^2 + g_{xx} dx^2 + g_{yy} dy^2 + g_{zz} dz^2 } along the path. In Dalespam's example only the x-coordinate varies so dt=dy=dz=0 along the path, meaning if the path varies from X1 to X2 and the path is timelike, you can calculate the proper time using the integral \int_{X1}^{X2} \sqrt{g_{xx} } \, dx. Does this answer your question "I mean to sayhow do you get the proper time interval to carry out the differentiation?"
The method prescribed in the quoted text indicates that the proper speed is equal to the speed of light.What speed should be observed by an observer standing on the ground say at the initial point of the motion?What time of travel should the stationary observer [say one at the initial point] calculate/observe?
 
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  • #67
Anamitra said:
The method prescribed in the quoted text indicates that the proper speed is equal to the speed of light.
How are you defining "proper speed"? The method given in my quote dealt only with calculating proper time, not "proper speed". Proper velocity is defined as the rate that coordinate position is changing relative to proper time (as opposed to coordinate time as with coordinate velocity), but it can take any value from 0 to infinity, so if you think proper speed would be "equal to the speed of light" I guess you have invented your own definition?
Anamitra said:
What speed should be observed by an observer standing on the ground say at the initial point of the motion?What time of travel should the stationary observer [say one at the initial point] calculate/observe?
Are you asking about coordinate time and speed in some coordinate system? (If so, which one? Is it the inertial rest frame of the observer, or is it some alternate coordinate system like the one DaleSpam discussed where a timelike worldline may have dt=0? If dt=0 for a journey in some coordinate system, then of course in that coordinate system the coordinate time of travel is zero and the coordinate speed is infinite)
 
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  • #68
The observer is at rest at the initial point of motion.He is trying to investigate the motion from a laboratory.

1) What formula should he use to calculate the speed or to define the speed of the particle at some point of the path?
2) What formula should he use to calculate the time interval or to define the time interval of the particle for the entire motion?
 
  • #70
Anamitra, given some timelike path, x, parameterized by a variable, lambda, the proper time along that path is defined* as:
\tau = \int \sqrt{-\frac{1}{c^2}\left(g_{\mu\nu} \frac{dx^{\mu}}{d\lambda} \frac{dx^{\nu}}{d\lambda} \right)} d\lambda

Since by definition the interval squared is always negative for a timelike interval, this value is always strictly positive, regardless of whether or not the coordinates are unusual.

*using the usual convention where spacelike intervals squared are positive and have units of length squared.
 
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  • #71
Anamitra said:
The observer is at rest at the initial point of motion.He is trying to investigate the motion from a laboratory.
In a non-stationary spacetime the term "at rest" has no clear meaning.
 
  • #72
DaleSpam said:
In a non-stationary spacetime the term "at rest" has no clear meaning.
The spatial distance between a pair of spatial points may change due to changes in g(11),g(22) and g(33) with time. But the coordinate labels[spatial] should not change. The physical distance between a pair of laboratories[along some spatial curve] may change with time but each laboratory should stand on the same coordinate labels relating to x,y and z. The laboratory is at rest refers to these unchanging coordinate labels.
This acknowledges the fact that the physical distance from some other spatial point[given by coordinate labels--x,y,z] may change with time.Calculation along some spatial curve is implied.
 
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  • #73
Anamitra said:
The spatial distance between a pair of spatial points may change due to changes in g(11),g(22) and g(33) with time. But the coordinate labels[spatial] should not change. The physical distance between a pair of laboratories[along some spatial curve] may change with time but each laboratory should stand on the same coordinate labels relating to x,y and z. The laboratory is at rest refers to these unchanging coordinate labels.
This acknowledges the fact that the physical distance from some other spatial point[given by coordinate labels--x,y,z] may change with time.Calculation along some spatial curve is implied.

I will describe a possibly relevant confusion I had that was clarified by Bcrowell and Dalespam in another thread a couple of months ago.

It is simply wrong to attach any physical meaning to the coordinate labels of an *arbitrary* coordinate system. They could be called a,b,c,d instead of z,y,t,z. Further, it is wrong to assume that the basis vector along some coordinate label at some spacetime point has the same character (timelike, spacelike, null) as at a different spacetime point. Bcrowell provided an example of a frequently used coordinate system for SR that has two lightlike basis vectors and two spacelike basis vectors and *no* timelike basis vectors. Yet it is not only well defined, it was particularly favored by Dirac (I think).

In this thread, early on, you (Anamitra) proposed a metric form where all the metric components included functional dependence on the coordinate named t as well as on e.g. x,y,z. Such a metric form almost necessarily would have the feature that the character of a given coordinate direction is different in different regions of spacetime.

Of course, for many purposes, you do want 'intuitive' coordinate systems. For a given geometry, you would build this up by choosing orthonormal geodesics from some point of interest (one timelike, 3 spacelike). Unfortunately, for general geometry, the larger the region you consider, the less Lorentzian the single coordinate system would be over the whole region.
 
  • #74
PAllen said:
I will describe a possibly relevant confusion I had that was clarified by Bcrowell and Dalespam in another thread a couple of months ago.

It is simply wrong to attach any physical meaning to the coordinate labels of an *arbitrary* coordinate system. They could be called a,b,c,d instead of z,y,t,z. Further, it is wrong to assume that the basis vector along some coordinate label at some spacetime point has the same character (timelike, spacelike, null) as at a different spacetime point. Bcrowell provided an example of a frequently used coordinate system for SR that has two lightlike basis vectors and two spacelike basis vectors and *no* timelike basis vectors. Yet it is not only well defined, it was particularly favored by Dirac (I think).

For the two coordinate systems we can have a one to one correspondence between the corresponding points(a,b,c,d) and (t,x,y,z) producing a consistent physical picture.We must have some transformation rule.One may try to make each of a,b,c and d dependent on coordinate time.The points will change position .
In our system I mean the(t,x,y,z) system--x,y,z are not changing with time

In the transformed system [(a,b,c,d) ],all the quantities can change with time if the transformation rule dictates such a condition[of time dependence]. The points should change position in a manner consistent with the gravitational effects

What ever the case is [When one thinks of different coordinate systems] we should be able to identify separately
1) Motion due to field.
2) Other types of motion

The distance between my house and the church at the end of the town along a road may go on changing with due to gravity. But to be aware of the motion wrt the road when I am driving to the church I must know the motion other than that due to the field.If such motion is not there I am at rest.
 
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  • #75
JesseM said:
Why do you say this path is "timelike"? In most coordinate systems typically used in physics (Minkowski coordinates, for example), if dt=0 along a path then the path is spacelike, not timelike.

Normally if dt=0 then the path is spacelike, but the actual definition for a spacelike path is that dS < 0 using the +--- signature. I think Anamitra is trying to make the case that there are certain conditions where dS^2 < 0 when dt=0.

For example below the event horizon of a Schwarzschild black hole, it is the sign of dS that defines the spacelike or timelike nature of the path rather than the value of dt and some people like to say that spacelike and timelike swap roles below the event horizon (informally speaking).

Just to clarify, in the Schwarzschild metric the interval along a radial path with no angular motion is:

dS^2 = (1-2m/r)dt^2 - \frac{dr^2}{(1-2m/r)}

When dt=0, ds^2 is negative when r>2m and positive when r<2m.

Above the event horizon where we can and have carried out actual experiments, a spacelike interval always has g_{tt}dt^2< g_{rr}dr^2 and dS^2<0. Below the event horizon, whether or the interval is spacelike depends upon whether you define spacelike as dt^2<dr^2 OR dS^2<0, because the two definitions do not agree here and we do not have (and probably never will have) any direct experimental measurements from that region. However, the conventional view is that the interval is timelike below the event horizon when g_{tt}dt^2< g_{rr}dr^2.

Now the question is what is the physical meaning of "timelike" beyond the mathematical definition of dS^2 > 0 ? One definition is that timelike events can be causally connected (while spacelike events cannot). Another observation is that while we can can move forwards or backwards along a spacelike path we can only move forwards (from the past to the future) along a timelike path outside the event horizon. Below the event horizon, the conventional interpretation is that we can move forwards or backwards relative to coordinate time but can only move in one direction relative to coordinate space (from the event horizon towards the central singularity). In other words a more general definition of a timelike path might be that "a timelike path is one in which you can only move in one direction relative to coordinate time or space". Note that defining timelike paths as ones that are contained within past and future light cones is a tautology, because the orientation of light cones depends on how you define timelike and spacelike in the first place.
 
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  • #76
Anamitra said:
I would request the audience to consider the following link in regard of the general nature of the ongoing conversation::

https://www.physicsforums.com/showpost.php?p=3062372&postcount=59

Anamitra said:
The time component of the four velocity[As suggested by PAllen] is zero. Further differentiation wrt to propertime[ds] yields time component of the momentum vector[multiplication by rest mass is required] which is again zero. It is zero energy particle![Assuming a particle is capable of moving along a timelike curve/path]

Could you make it clear what you mean by the "time component of the four velocity". Let us define the four velocity as:

\frac{c^2dt^2}{d\tau^2} - \frac{dx^2}{d\tau^2} - \frac{dy^2}{d\tau^2} - \frac{dz^2}{d\tau^2} = (x_0, x_1, x_2, x_3)

This quantity is always equal to the speed of light squared.

I would normally take the time component to mean x_0. Is that what you mean? When this component has the value zero, the four velocity is still the speed of light (squared). Further differentiation would give the four acceleration and a value of zero here just means that the particle is inertial with constant velocity, which does not by itself imply zero momentum or energy.

Now let us take a look at the "nature of dS" at least in the context of Minkowski space before moving onto GR. Here dS is defined two dimensionally by:

dS^2 = c^2d\tau^2 = c^2 dt^2 - dr^2

Using units of c=1, dS is essentially the proper time measured by a particle that travels a coordinate distance dx in a coordinate time dt. For a spacelike interval dS turns out to be imaginary. This imaginary quantity is the (imaginary) proper time measured by an (imaginary) particle traveling at greater than the speed of light (such as a hypothetical tachyon), because no real particle with positive proper mass can travel a distance dx in less than time dt. While the velocity of this imaginary particle appears to be infinite in the original frame, transformation to other reference frames can yield a finite velocity for the imaginary particle (but still greater than the speed of light) or a greater than infinite velocity (!) going backwards in time.
 
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  • #77
yuiop said:
Note that defining timelike paths as ones that are contained within past and future light cones is a tautology, because the orientation of light cones depends on how you define timelike and spacelike in the first place.

I would only partially agree with this. Null cones are completely defined by the geometry, and some authors use them to characterize all essential features of the geometry. Which you pick as past and future is arbitrary, and making such a choice is a step towards introducing a coordinate system. As to inside / outside, I agree that I don't know of any definition other than the family of paths along which ds^2 is positive (if you use (+---) signature or negative if you use (-+++) signature. I wouldn't be surpised if there were some more fundamental defintion of inside, but I haven't run across it.
 
  • #78
Anamitra said:
1) What formula should he use to calculate the speed or to define the speed of the particle at some point of the path?
In most situations a physicist would just use the coordinate speed I think, but there are other options, for example in cosmology the "velocity" in the Hubble formula is based on the rate at which proper distance (measured along a path confined to a single surface of simultaneity in cosmological coordinates) changes with coordinate time, and as I mentioned above "proper velocity" measures the rate at which coordinate distance changes with the moving object's own proper time. If you want a specific formula you have to specify what notion of "speed" you want to measure.
Anamitra said:
2) What formula should he use to calculate the time interval or to define the time interval of the particle for the entire motion?
What do you mean by "time interval"? The coordinate time in his frame? The particle's own proper time? In the case of proper time I already gave you the formula for that...
 
  • #79
I think the speed of a particle measured by a 'scientist' can be given in a coordinate independent way. The particle has some 4 velocity at some event on its world line; assume the 'scientist' has some different 4 velocity at the same event. The scientist would define the time axis to be their 4 velocity; they could define a spacelike unit vector normal (in the spacetime sense) to their 4 velocity in the direction of the particle's spatial motion. The the particle's 4 velocity dot product this spatial vector, divided by the particle's 4 velocity dot the scientist's 4 velocity, would be the particle's speed as perceived by the scientist.

Note that even for as simple a case Swarzschild coordinates, you wouldn't directly use the coordinates to define intuitive speed of anything - two of them are angles..
 
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  • #80
Anamitra said:
The spatial distance between a pair of spatial points may change due to changes in g(11),g(22) and g(33) with time. But the coordinate labels[spatial] should not change.
That is just the point. The coordinate labels may very well change. Simply because a coordinate is labeled "t" does not imply that it is timelike. There are many such examples.

1) in the maximally extended Schwarzschild solution in the interior region the coordinate labeled r is timelike and the coordinate labeled t is spacelike.

2) in rotating coordinates in flat spacetime the coordinate labeled t becomes spacelike for r &gt; \frac{c}{2 \pi \omega}

3) there are many coordinate systems possible where the coordinates are not orthonormal, in such systems you may have null coordinates, or more than one timelike coordinate, etc.

You cannot rely on the label of the coordinate, nor even on its position within the list of coordinates, to tell you what the coordinate means physically. The coordinates themselves are almost completely arbitrary labels, like addresses or zip codes. It is the metric which gives them physical meaning, determining if they are orthogonal, normal, constant, timelike, spacelike, etc.
 
  • #81
yuiop said:
Could you make it clear what you mean by the "time component of the four velocity". Let us define the four velocity as:

\frac{c^2dt^2}{d\tau^2} - \frac{dx^2}{d\tau^2} - \frac{dy^2}{d\tau^2} - \frac{dz^2}{d\tau^2} = (x_0, x_1, x_2, x_3)

This quantity is always equal to the speed of light squared.

I would normally take the time component to mean x_0. Is that what you mean? When this component has the value zero, the four velocity is still the speed of light (squared). Further differentiation would give the four acceleration and a value of zero here just means that the particle is inertial with constant velocity, which does not by itself imply zero momentum or energy.

.

Actually the further differentiation of four velocity is not required.[there was some inadvertence from my side there]The time component of the momentum vector is energy.If {dt}{/}{{d}{{\tau}}{=}{0} ,energy is zero[the time component of the momentum vector is zero in this case]. Further differentiation is not required.

[One should use the metric coefficients in the formula represented in the quoted text]
 
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  • #82
JesseM said:
In most situations a physicist would just use the coordinate speed I think, but there are other options, for example in cosmology the "velocity" in the Hubble formula is based on the rate at which proper distance (measured along a path confined to a single surface of simultaneity in cosmological coordinates) changes with coordinate time, and as I mentioned above "proper velocity" measures the rate at which coordinate distance changes with the moving object's own proper time. If you want a specific formula you have to specify what notion of "speed" you want to measure.

The coordinate speed of light is not constant[in vacuum]. The physical speed is constant.For the evaluation of physical speed, one has to consider physical time.What alternative could you offer?

JesseM said:
What do you mean by "time interval"? The coordinate time in his frame? The particle's own proper time? In the case of proper time I already gave you the formula for that...

If a person sees a particle flying out of his laboratory he should not use the concept of proper time[if he observes the particle getting absorbed somewhere out in space and he wants to have an estimate of the time interval].For calculating time intervals for events occurring in the laboratory at some specified location[inside the laboratory], he uses the formula

dT=g(00)dt

He observes the particle getting absorbed somewhere out in space.Is coordinate time interval going to be a proper estimate of the time elapsed?
 
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  • #83
PAllen said:
I think the speed of a particle measured by a 'scientist' can be given in a coordinate independent way. The particle has some 4 velocity at some event on its world line; assume the 'scientist' has some different 4 velocity at the same event. The scientist would define the time axis to be their 4 velocity; they could define a spacelike unit vector normal (in the spacetime sense) to their 4 velocity in the direction of the particle's spatial motion. The the particle's 4 velocity dot product this spatial vector, divided by the particle's 4 velocity dot the scientist's 4 velocity, would be the particle's speed as perceived by the scientist.

Note that even for as simple a case Swarzschild coordinates, you wouldn't directly use the coordinates to define intuitive speed of anything - two of them are angles..

In my system --the traditional [t,x,y,z]rectangular system I am getting the time component of the momentum four-velocity--as zero[in some particular situation]. Consequently energy works out to be zero.

If you get the value of energy as a non zero quantity in your frame/system of coordinates, how do you interpret the situation----as something physically different from what I am getting?
 
  • #84
In relation to my previous post:

Energy is a conserved quantity but not an invariant.But the rest energy part should not change.
 
  • #85
JesseM said:
Coordinate time is completely irrelevant, only proper time matters if you want to know what physical clocks are doing. If you think it somehow makes a difference that dt=0, I think you're misunderstanding something basic about the physical meaning of proper time (and the fact that it's independent of what coordinate system you choose--we could easily describe the exact same physical path in a different coordinate system where dt was not zero, and the proper time along the path would necessarily be exactly the same).

A particle flies out of my laboratory and gets absorbed somewhere out in space.Do I have any notion of the time elapsed between the events--the particle flying out and getting absorbed somewhere out in space? Can we use the concept of proper time here?

[NB:For events happening inside the laboratory[at a specified location inside the laboratory] one does not use coordinate time.One should not consider its use logical in this case also--the particle flying out of the laboratory and getting absorbed.Coordinate time does not take into account the individual clock rates of the different points ]
 
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  • #86
Anamitra said:
In my system --the traditional [t,x,y,z]rectangular system I am getting the time component of the momentum four-velocity--as zero[in some particular situation]. Consequently energy works out to be zero.

If you get the value of energy as a non zero quantity in your frame/system of coordinates, how do you interpret the situation----as something physically different from what I am getting?

You keep insisting on say energy is the component labeled 't'. The universe knows about our letters? If coordinate is not timelike in some region of spacetime, its corresponding vector component is unrelated to energy. The energy measured by an observer defined in a coordinate independent way is the dot product of the particle's 4 momentum with the observer's 4 velocity.
 
  • #87
Anamitra said:
In relation to my previous post:

Energy is a conserved quantity but not an invariant.But the rest energy part should not change.

Rest energy is the norm of the 4 momentum, which is, indeed coordinate independent and is defined based on the metric and the 4 momentum. It is *not* defined in terms the time component of 4 momentum unless you are using locally Lorentz coordinates. Since early in this thread, you proposed a dynamic metric, this means you would have to re-specify a coordinate mapping in the vicinity of each event of interest (if you want to ensure they are locally Lorentz). You can't start with some coordinates locally Lorentz at one event and pretend they are such at a different event.
 
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  • #88
I think Anamitra may have orginally had in mind the idea of a coordinate path like x=t/2 for t=(1,2) versus t=(5,6), with one being spacelike and one being timelike, and wasn't so mystified by this. Dalespam and I just simplified this to the case of curve of constant t having different character at different regions of spacetime. To us, this is not strange or difficult at all, just a simpler case of the same thing (take my guess at Anamitra's case; perform trivial coordinate tranform, and you end up with the simpler case Dalespam and I were discussing). It seems our attempt to simplify greatly confused Anamitra. Ultimately, it should be very helpful for Anamitra to understand the stripped down case.
 
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  • #89
Let us first come back to the basic issue---whether there there can be an intercoversion between a spacelike path and a timelike/null path by the effect of gravity[Due to changes in the values/expressions representing the metric coefficients in a time dependent field]
{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}

The value of the metric coefficients can always change by the action of gravity---this is a known fact.The sign of ds^2 can also change[I am referring to such a possibility here] . By choosing a different coordinate system you can never change the physical consequences.

Even if you consider some weird geometry with unusual signs for the metric coefficients[in case such an action is possible]or even if you apply any other type of contrivance, the huge number of cases[in relation to the interconversion] indicated in the last paragraph cannot be ruled out.

The light cone mechanism has been clearly depicted in the following link:
https://www.physicsforums.com/showpost.php?p=3061481&postcount=38

Many other side issues have come up in this thread.I will definitely address them[and I have been addressing them]
 
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  • #90
Anamitra said:
Let us first come back to the basic issue---whether there there can be an intercoversion between a spacelike path and a timelike/null path by the effect of gravity
{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}

Let's try a particular example in Schwarzschild metric. Assume dx2 and dx3 are zero so we are considering the two dimensional radial case so:

ds^2 = (1-2M/r)dt^2 - (1-2M/r)^{-1} dr^2

For a timelike path, ds^2 is positive. Below the event horizon (say r=M) a path with dt=0 is a valid timelike path because:

ds^2 = - (1-2M/M)^{-1} dr^2 = +dr^2

Can we calculate a permissible velocity, momentum and energy of such a particle?
 
  • #91
You don't have to go below the event horizon. Please stay above it and consider changes in the coefficients:
{{(}{1}{-}{2M}{/}{r}{)}} and {{(}{1}{-}{2M}{/}{r}{)}}^{-1}


Such changes are quite possible if some high density erratic mass distribution comes near the mass M. The resultant metric will be quite different[the coefficients will undergo a huge change in their functional expression].A space like path may get converted into a timelike one and vice versa. [It is not necessary to make dt=0 for such considerations]

[The time dependence of a gravitational field is related to the changes in the values/expressions of the metric coefficients]
 
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  • #92
Regarding Time dependence:I can always choose a frame of reference where the spatial coordinate labels [x,y,z] do not change with time for my inferences.The values/expressions for the metrics do change due to gravitational effects.PAllen and others can always choose a frame where the coordinates are changing. The physical nature of the conclusions should not change.

A simple illustration:
I am in a laboratory at A(x1,y1,z1) . Pallen and yuiop are at different one B(x2,y2,z2). A few minutes ago we had spacelike paths between the two labs. Now we have timelike paths thaks to the gravitational effects.

Would the physical nature of the conclusions change if by some suitable transformation the spatial coordinates are made to vary with time?
 
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  • #93
Anamitra said:
Let us first come back to the basic issue---whether there there can be an intercoversion between a spacelike path and a timelike/null path by the effect of gravity[Due to changes in the values/expressions representing the metric coefficients in a time dependent field]
{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}
This is a strange way to word it. A given spacetime path is, by definition, fixed in time and character. I think what you must mean is that a similar (e.g same coordinate slope) coordinate path in different regions of spacetime can be spacelike in one region and timelike in another. That is obviously true, and whenever it is true, a trivial coordinate transform can convert it to the simpler case of a coordinate axis having different character in different regions of spacetime.
Anamitra said:
The value of the metric coefficients can always change by the action of gravity---this is a known fact.The sign of ds^2 can also change[I am referring to such a possibility here] . By choosing a different coordinate system you can never change the physical consequences.
Correct. Change in coordinate system will never change an invariant like the spacelike/timelike character of a path.
Anamitra said:
Even if you consider some weird geometry with unusual signs for the metric coefficients[in case such an action is possible]or even if you apply any other type of contrivance, the huge number of cases[in relation to the interconversion] indicated in the last paragraph cannot be ruled out.
Of course. No one disagreed with this. From my point of view, I was simply proposing the simplest example of this.
Anamitra said:
The light cone mechanism has been clearly depicted in the following link:
https://www.physicsforums.com/showpost.php?p=3061481&postcount=38

Many other side issues have come up in this thread.I will definitely address them[and I have been addressing them]

Actually, they are mostly all the same issue. You think they are different for superficial reasons (like whether dt=0 or not).
 
  • #94
Anamitra said:
Let us first come back to the basic issue---whether there there can be an intercoversion between a spacelike path and a timelike/null path by the effect of gravity
The answer to that is very clearly and definitively, "no".

What can happen is that there may be two different paths which differ only by a constant offset of one of the coordinates (e.g. t) and one of these two different paths may be spacelike while the other is timelike.
 
  • #95
Anamitra said:
A simple illustration:
I am in a laboratory at A(x1,y1,z1) . Pallen and yuiop are at different one B(x2,y2,z2). A few minutes ago we had spacelike paths between the two labs. Now we have timelike paths thaks to the gravitational effects.
If you let y1=y2=0 and z1=z2=0 then this is exactly the example I gave previously. Note that the spacelike and timelike paths are different paths because the t coordinate differs by a few minutes.
 
  • #96
Anamitra said:
Regarding Time dependence:I can always choose a frame of reference where the spatial coordinate labels [x,y,z] do not change with time for my inferences.The values/expressions for the metrics do change due to gravitational effects.PAllen and others can always choose a frame where the coordinates are changing. The physical nature of the conclusions should not change.

A simple illustration:
I am in a laboratory at A(x1,y1,z1) . Pallen and yuiop are at different one B(x2,y2,z2). A few minutes ago we had spacelike paths between the two labs. Now we have timelike paths thaks to the gravitational effects.

Would the physical nature of the conclusions change if by some suitable transformation the spatial coordinates are made to vary with time?

This is getting at your confusion. You can speak of a space *time* path being spacelike. What you mean is that the path between:

(t1,x1,y1,z1) and (t1,x2,y2,z2) maintaining t=t1 is spacelike,

while the path:

(t2,x1,y1,z1) and (t2,x2,y2,z2) maintaining t=t2 is timelike.

These are two completely independent paths through spacetime, and the situation implies, mostly, that the meaning of the coordinates *has* changed. There is, presumably, a family of spacelike paths connecting the world lines of the two labs at different proper time points along the world lines. The situation above simply means the the coordinate representation of these paths looks very different at different points along the worldlines.
 
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  • #97
Anamitra said:
Let us first come back to the basic issue---whether there there can be an intercoversion between a spacelike path and a timelike/null path by the effect of gravity[Due to changes in the values/expressions representing the metric coefficients in a time dependent field]
{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}
PAllen said:
This is a strange way to word it. A given spacetime path is, by definition, fixed in time and character. I think what you must mean is that a similar (e.g same coordinate slope) coordinate path in different regions of spacetime can be spacelike in one region and timelike in another. That is obviously true, and whenever it is true, a trivial coordinate transform can convert it to the simpler case of a coordinate axis having different character in different regions of spacetime.
A spacetime path remains spacetime in time independent gravitational fields. Their nature remains invariant wrt coordinate transformation[in all types of fields time dependent or independent].In my illustration the coordinate system is not being changed or transformed.We are considering changes in the metric coefficients in the same coordinate system(t,x,y,z). Conclusions should remain unchanged for all other frames once the change [in tne metric coefficients]has taken place.
PAllen said:
Correct. Change in coordinate system will never change an invariant like the spacelike/timelike character of a path.
This does not come in the way of my arguments
 
  • #98
yuiop said:
Let's try a particular example in Schwarzschild metric. Assume dx2 and dx3 are zero so we are considering the two dimensional radial case so:

ds^2 = (1-2M/r)dt^2 - (1-2M/r)^{-1} dr^2

For a timelike path, ds^2 is positive. Below the event horizon (say r=M) a path with dt=0 is a valid timelike path because:

ds^2 = - (1-2M/M)^{-1} dr^2 = +dr^2

Can we calculate a permissible velocity, momentum and energy of such a particle?

Of course. You need to specifiy a path. If you say dt=0 and r=M, you have an event not a path. You can say, e.g. r varies from M to .9M, while t=t0. At r=M, you get a 4 velocity of (vt,vr)=(0,1), a contravariant momentum of (0,m); and if you take the norm of the momentum using the metric, you get m as rest energy (of course).
 
  • #99
PAllen said:
This is getting at your confusion. You can speak of a space *time* path being spacelike. What you mean si that the path between:

(t1,x1,y1,z1) and (t1,x2,y2,z2) maintaining t=t1 is spacelike,

while the path:

(t2,x1,y1,z1) and (t2,x2,y2,z2) maintaining t=t2 is timelike.

These are two completely independent paths through spacetime, and the situation implies, mostly, that the meaning of the coordinates *has* changed. There is, presumably, a family of spacelike paths connecting the world lines of the two labs at different proper time points along the world lines. The situation above simply means the the coordinate representation of these paths looks very different at different points along the worldlines.

This part is absolutely OK. Thanks for that PAllen!
Now we are thinking of two spacetime paths A and B where B is a subset of A.t for initial point of A is greater than t for initial point of B. The part between initial point of A and initial point of B is time like and the rest is spacelike when the observer is at A.When the observer reaches the initial point of B he is amazed to find that the rest of the journey can be carried out since the remailing path has become timelike due to gravity!

The example in the following link is in tune with what you are saying!
https://www.physicsforums.com/showpost.php?p=3061386&postcount=36
 
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  • #100
Anamitra said:
This part is absolutely OK. Thanks for that PAllen!
Now we are thinking of two spacetime paths A and B where B is a subset of A.t for initial point of A is greater than t for initial point of B. The part between initial point of A and initial point of B is time like and the rest is spacelike when the observer is at A.When the observer reaches the initial point of B he is amazed to find that the rest of the journey can be carried out since the path has become timelike!

The example in the following link is in tune with what you are saying!
https://www.physicsforums.com/showpost.php?p=3061386&postcount=36

Sorry, but this doesn't really make sense. The path from beginning of A to B cannot be the the path of any observer through spacetime. This path is truly like the following (and can be converted to it by coordinate transform):

imagine a born rigid ruler at some t=t0; this ruler represents the beginning of A to beginning of B; Now imagine the worldline of some observer that intersects the ruler at the beginning of B. This is the rest of this mixed spacetime path. You are simply drawing a path through spacetime that follows a ruler, then follows a worldline. This is not a very meaningful path, but you can define it.

[Edit: I accidentally reversed which part of the path is which from what you proposed, but the idea is the same see my later reply for additional point]
 
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